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How To Solve An Initial Value Problem (5 Key Steps To Take)

Initial value problems come up often in calculus, physics, and other subjects.  You can solve some of them with straightforward antiderivatives, while others will require you to solve a challenging ordinary differential equation.

Of course, we can solve initial value problems in “layers”.  For example, in physics, we often start with acceleration (due to gravity) and find the height function for an object, given its starting height and speed.

Let’s get started.

What Is An Initial Value Problem?

An initial value problem (IVP) uses calculus, a differential equation, and a starting condition to find a function that models the problem.  For example, a common IVP in physics is to find an equation for the height of a falling object, given its starting height and velocity.

An initial value problem involves an ordinary differential equation (ODE).  An ordinary differential equation is made up of one or more functions and their derivatives .

Now that we know what an initial value problem is, let’s find out how to solve one.

How To Solve An Initial Value Problem

There are five key steps you can take to help you solve an initial value problem.

It will help to see these steps applied to a real problem, so let’s take a look at some now.

Example 1: How To Solve An Initial Value Problem

Let’s say that you are somewhere due south of Boston.  You start driving south at a constant speed of 60 miles per hour.

To solve this problem, we’ll take the 5 steps listed above.

Step 1: write out the equation.

The equation tells us the speed S of the vehicle at a given time t.  So, we have the equation:

This suggests that the speed of the car is a constant 60 (miles per hour).

Remember that speed is the derivative of position, meaning S(t) = P’(t).  So, we can rewrite the equation as:

We are given that the car is 220 miles south of Boston after 3.2 hours.  This implies a position of P = 220 miles at a time t = 3.2 hours.

As an ordered pair, we can write this as (3.2, 220).

Since the derivative P’(t) is already isolated in the equation P’(t) = 60, we don’t need to do anything here.

Step 5: use the starting condition.

Using these values in the equation we found in Step 4 gives us:

For example, at time t = 0, the car is P(0) = 60(0) + 28 = 28 miles south of Boston (that is, you started 28 miles south of Boston).

Example 2: How To Solve An Initial Value Problem

Let’s try a classic initial value problem from physics.  This one has two “layers”, and two antiderivatives must be taken: the first to go from acceleration to velocity, and the second to go from velocity to position.

A ball is thrown downward from the top of a building.  The starting velocity is -20 feet per second (the negative denotes that it is falling, or moving towards Earth).  The starting height is 500 feet.

We are not given any variables, so we will need our own.  Let’s use A for the acceleration of the ball, V for the velocity of the ball, P for the position (height) of the ball, and t for the time (in seconds).

This suggests that the acceleration of the ball is a constant -32 (feet per second, per second).  This is the acceleration due to gravity.

Step 2: identify the starting condition.

Step 3: isolate the derivative.

This is not too difficult, since the antiderivative of a constant c is ct.  Taking the antiderivative on both sides of our equation gives us:

where K is an unknown constant.

However, we’re not done yet.  We want to find the position function, so we have to go through the steps again (this time, to find position from velocity).

The equation we just found tells us the velocity V of the ball at a given time t.  So, we have the equation:

This is not too difficult, since the antiderivative of a constant c is ct, and the antiderivative of a linear term bt is bt 2 /2.  Taking the antiderivative on both sides of our equation gives us:

Now, we use the starting condition to solve for the constant K.  Remember from Step 2 that our starting condition was P = 500 at t = 0.

So, the entire equation is P(t) = -16t 2 – 20t + 500.  This tells us the position of the ball at t seconds.

The ball is at a height of 350 feet above ground at 2.5 seconds.

Velocity is used in other applications besides initial value problems – you can learn about what velocity is used for in this article.

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Solving Initial Value Problems (IVPs) A Comprehensive Guide

// Last Updated: April 17, 2023 - Watch Video //

Did you know that when you solve a differential equation with a specific condition, you’re tackling an initial value problem ?

Jenn, Founder Calcworkshop ® , 15+ Years Experience (Licensed & Certified Teacher)

In simpler terms, you’re looking for a solution that meets certain requirements to find a unique answer .

In a previous lesson, you learned about Ordinary Differential Equations . Now, you’ll dive deeper to explore n-parameter family of solutions and use an initial condition (IC) to figure out the constants in that family.

Understanding n-parameter Family of Solutions

So, what’s an n-parameter family?

Well, a first-order differential equation \(F\left(x, y, y^{\prime}\right)=0\) typically has a single arbitrary constant called a parameter \(\mathrm{c}\), whereas a second-order differential equation \(F\left(x, y, y^{\prime}, y^{\prime \prime}\right)=0\) usually has two arbitrary constants, denoted \(c_{1}\) and \(c_{2}\).

One Two Three Parameter —  Solution Formulas

Therefore, a single differential equation can possess an infinite number of solutions corresponding to the unlimited number of choices for the parameters.

So, an n-parameter family of solutions of a given nth-order differential equation represents the set of all solutions of the equation. And we typically refer to an \(n\)-parameter family of solutions as the general solution – it’s easier to say!

Consequently, if we assign arbitrary constants to the differential equation, referred to as initial conditions, our general solution is called a particular solution because it represents a specific answer for some particular constraint.

It should also be noted that solutions of an \(\mathrm{n}\)-th order differential equation that are not included in the general solution are called singular solutions.

Example: Verifying and Finding Solutions to Initial Value Problems

Let’s look at an example of how we will verify and find a solution to an initial value problem given an ordinary differential equation.

Verify that the function \(y=c_{1} e^{2 x}+c_{2} e^{-2 x}\) is a solution of the differential equation \(y^{\prime \prime}-4 y=0\).

Then find a solution of the second-order IVP consisting of the differential equation that satisfies the initial conditions \(y(0)=1\) and \(y^{\prime}(0)=2\).

First, we will verify that the function is a solution by noticing that we are given a two-parameter family of solutions because we have a second-order differential equation. Therefore, we need to find the second derivative of our function.

\begin{align*} \begin{aligned} & y=c_{1} e^{2 x}+c_{2} e^{-2 x} \\ & y^{\prime}=2 c_{1} e^{2 x}-2 c_{2} e^{-2 x} \\ & y^{\prime \prime}=4 c_{1} e^{2 x}+4 c_{2} e^{-2 x} \end{aligned} \end{align*}

Now, we will substitute our derivatives into the ODE and verify that the left-hand side equals the right-hand side.

\begin{equation} \begin{aligned} & y^{\prime \prime}-4 y=0 \\ & \left(4 c_1 e^{2 x}+4 c_2 e^{-2 x}\right)-4\left(c_1 e^{2 x}+c_2 e^{-2 x}\right)=0 \\ & 4 c_1 e^{2 x}+4 c_2 e^{-2 x}-4 c_1 e^{2 x}-4 c_2 e^{-2 x}=0 \\ & 0=0 \end{aligned} \end{equation}

Now, we will find a solution to the second-order IVP by substituting the initial conditions into their corresponding functions.

\begin{equation} \text { If } y=c_1 e^{2 x}+c_2 e^{-2 x} \text { and } y(0)=1, \text { then } \end{equation}

\begin{align*} 1=c_{1} e^{2(0)}+c_{2} e^{-2(0)} \Rightarrow 1=c_{1}(1)+c_{2}(1) \Rightarrow 1=c_{1}+c_{2} \end{align*}

\begin{equation} \text { If } y^{\prime}=2 c_1 e^{2 x}-2 c_2 e^{-2 x} \text { and } y^{\prime}(0)=2 \text {, then } \end{equation}

\begin{align*} 2=2 c_{1} e^{2(0)}-2 c_{2} e^{-2(0)} \Rightarrow 2=2 c_{1}(1)-2(1) \Rightarrow 2=2 c_{1}-2 c_{2} \end{align*}

Next, we will solve the resulting system for \(c_{1}\) and \(c_{2}\).

\begin{align*} \left\{\begin{array} { c } { 1 = c _ { 1 } + c _ { 2 } } \\ { 2 = 2 c _ { 1 } – 2 c _ { 2 } } \end{array} \Rightarrow \left\{\begin{array}{c} c_{1}+c_{2}=1 \\ c_{1}-c_{2}=1 \end{array} \Rightarrow c_{1}=1 \quad \text { and } \quad c_{2}=0\right.\right. \end{align*}

Therefore, the particular solution for the IVP given the initial condition is:

\begin{equation} \begin{aligned} & y=(1) e^{2 x}+(0) e^{-2 x} \\ & y=e^{2 x} \end{aligned} \end{equation}

I find that it’s helpful to remember that Initial condition(s) are values of the solution and/or its derivative(s) at specific points. Which means, according to Paul’s Online Notes that solutions to “nice enough” differential equations are unique; hence, only one solution will meet the given conditions.

The Game-Changing Existence and Uniqueness Theorem

But how do we know there will be a solution to the differential equation?

The Existence of a Unique Solution Theorem is a key concept in this course. It provides specific conditions that ensure a unique solution exists for an Initial Value Problem (IVP).

Definition: The existence-unique solution theorem says that if we let \(\mathrm{R}\) be a rectangular region in the \(x y-\) plane defined by \(a \leq x \leq b, c \leq y \leq d\) that contains the point \(\left(x_{0}, y_{0}\right)\) in its interior. And if \(f(x, y)\) and \(\frac{\partial f}{\partial y}\) are continuous on \(\mathrm{R}\), then there exists some interval \(\left(x_{0}-h, x_{0}+h\right), h>0\), contained in \([a, b]\), and a unique function \(y(x)\), defined on \(I_{0}\), that is a solution of the initial value problem.

That’s pretty “mathy” right?!

In *sorta* simpler terms, the existence-unique solution theorem essentially states that under specific conditions, there is a unique solution for an initial value problem. If a point (x₀, y₀) is within a rectangular region R in the xy-plane, and both the function f(x, y) and its partial derivative with respect to y are continuous in R , then there is an interval (x₀-h, x₀+h), with h>0, contained within the range [a, b]. Within this interval, a unique function y(x) exists as a solution to the initial value problem.

Existence Uniqueness Theorem — Graphical

Example: Solving an IVP with Given Initial Conditions

Let’s break this down into easy-to-understand steps by working an example.

Determine whether the existence-uniqueness theorem implies the given initial value problem has a unique solution through the given point.

\begin{equation} y^{\prime}=y^{2 / 3},(8,4) \end{equation}

First, we will verify that our ODE is continuous by letting \(f(x, y)=y^{\prime}\) and graphing the curve.

\begin{equation} f(x, y)=y^{2 / 3} \end{equation}

ODE — 3D Graph

So, \(f(x, y)\) is continuous for all real numbers.

Next, we will take the partial derivative with respect to \(\mathrm{y}\) and determine if the partial derivative is also continuous.

\begin{align*} \frac{\partial f}{\partial y}=f_{y}=\frac{2}{3} y^{-1 / 3}=\frac{2}{3}\left(\frac{1}{\sqrt[3]{y}}\right) \end{align*}

This indicates that a unique solution exists when \(y>0\).

Therefore, we can safely conclude that our given point \((8,4)\) will provide a unique solution because our \(y\)-value is greater than zero.

Going forward…

So, together we will dive into the world of n-th parameter family of solutions, find solutions for initial value problems, and determine the existence of a solution and whether a differential equation contains a unique solution through a given point.

Let’s jump right in.

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Module 4: Applications of Derivatives

Initial-value problems, learning outcomes.

• Use antidifferentiation to solve simple initial-value problems

We look at techniques for integrating a large variety of functions involving products, quotients, and compositions later in the text. Here we turn to one common use for antiderivatives that arises often in many applications: solving differential equations.

A differential equation is an equation that relates an unknown function and one or more of its derivatives. The equation

is a simple example of a differential equation. Solving this equation means finding a function [latex]y[/latex] with a derivative [latex]f[/latex]. Therefore, the solutions of [latex]\frac{dy}{dx}[/latex] are the antiderivatives of [latex]f[/latex]. If [latex]F[/latex] is one antiderivative of [latex]f[/latex], every function of the form [latex]y=F(x)+C[/latex] is a solution of that differential equation. For example, the solutions of

are given by

Sometimes we are interested in determining whether a particular solution curve passes through a certain point [latex](x_0,y_0)[/latex]—that is, [latex]y(x_0)=y_0[/latex]. The problem of finding a function [latex]y[/latex] that satisfies a differential equation

with the additional condition

is an example of an initial-value problem . The condition [latex]y(x_0)=y_0[/latex] is known as an initial condition . For example, looking for a function [latex]y[/latex] that satisfies the differential equation

and the initial condition

is an example of an initial-value problem. Since the solutions of the differential equation are [latex]y=2x^3+C[/latex], to find a function [latex]y[/latex] that also satisfies the initial condition, we need to find [latex]C[/latex] such that [latex]y(1)=2(1)^3+C=5[/latex]. From this equation, we see that [latex]C=3[/latex], and we conclude that [latex]y=2x^3+3[/latex] is the solution of this initial-value problem as shown in the following graph.

Figure 2. Some of the solution curves of the differential equation [latex]\frac{dy}{dx}=6x^2[/latex] are displayed. The function [latex]y=2x^3+3[/latex] satisfies the differential equation and the initial condition [latex]y(1)=5[/latex].

Example: Solving an Initial-Value Problem

Solve the initial-value problem

First we need to solve the differential equation. If [latex]\frac{dy}{dx}= \sin x,[/latex] then

Next we need to look for a solution [latex]y[/latex] that satisfies the initial condition. The initial condition [latex]y(0)=5[/latex] means we need a constant [latex]C[/latex] such that [latex]− \cos x+C=5[/latex]. Therefore,

The solution of the initial-value problem is [latex]y=− \cos x+6[/latex].

Solve the initial value problem [latex]\frac{dy}{dx}=3x^{-2}, \,\,\, y(1)=2[/latex].

Find all antiderivatives of [latex]f(x)=3x^{-2}[/latex].

[latex]y=-\frac{3}{x}+5[/latex]

Watch the following video to see the worked solution to Example: Solving an Initial-Value Problem and the above Try It.

Initial-value problems arise in many applications. Next we consider a problem in which a driver applies the brakes in a car. We are interested in how long it takes for the car to stop. Recall that the velocity function [latex]v(t)[/latex] is the derivative of a position function [latex]s(t)[/latex], and the acceleration [latex]a(t)[/latex] is the derivative of the velocity function. In earlier examples in the text, we could calculate the velocity from the position and then compute the acceleration from the velocity. In the next example, we work the other way around. Given an acceleration function, we calculate the velocity function. We then use the velocity function to determine the position function.

Example: Decelerating Car

A car is traveling at the rate of [latex]88[/latex] ft/sec ([latex]60[/latex] mph) when the brakes are applied. The car begins decelerating at a constant rate of [latex]15[/latex] ft/sec 2 .

• How many seconds elapse before the car stops?
• How far does the car travel during that time?

The acceleration is the derivative of the velocity,

Therefore, we have an initial-value problem to solve:

Integrating, we find that

Since [latex]v(0)=88, \, C=88[/latex]. Thus, the velocity function is

Integrating, we have

Since [latex]s(0)=0[/latex], the constant is [latex]C=0[/latex]. Therefore, the position function is

Watch the following video to see the worked solution to Example: Decelerating Car.

Suppose the car is traveling at the rate of 44 ft/sec. How long does it take for the car to stop? How far will the car travel?

[latex]v(t)=-15t+44[/latex]

2.93 sec, 64.5 ft

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Solve Initial Value Problem-Definition, Application and Examples

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Existence and Uniqueness

Continuity and differentiability, dependence on initial conditions, local vs. global solutions, higher order odes, boundary behavior, particular and general solutions, engineering, biology and medicine, economics and finance, environmental science, computer science, control systems.

Solving initial value problems (IVPs) is an important concept in differential equations . Like the unique key that opens a specific door, an initial condition can unlock a unique solution to a differential equation.

As we dive into this article, we aim to unravel the mysterious process of solving initial value problems in differential equations . This article offers an immersive experience to newcomers intrigued by calculus’s wonders and experienced mathematicians looking for a comprehensive refresher.

Solving the Initial Value Problem 

To solve an initial value problem , integrate the given differential equation to find the general solution. Then, use the initial conditions provided to determine the specific constants of integration.

An initial value problem (IVP) is a specific problem in differential equations . Here is the formal definition. An initial value problem is a differential equation with a specified value of the unknown function at a given point in the domain of the solution.

More concretely, an initial value problem is typically written in the following form:

dy/dt = f(t, y) with y(t₀) = y₀

• dy/dt = f(t, y) is the differential equation , which describes the rate of change of the function y with respect to the variable t .
• t₀ is the given point in the domain , often time in many physical problems .
• y(t₀) = y₀ is the initial condition , which specifies the value of the function y at the point t₀.

An initial value problem aims to find the function y(t) that satisfies both the differential equation and the initial condition . The solution y(t) to the IVP is not just any solution to the differential equation , but specifically, the one which passes through the point (t₀, y₀) on the (t, y) plane.

Because the solution of a differential equation is a family of functions, the initial condition is used to find the particular solution that satisfies this condition. This differentiates an initial value problem from a boundary value problem , where conditions are specified at multiple points or boundaries.

Solve the IVP y’ = 1 + y^2, y(0) = 0 .

This is a standard form of a first-order non-linear differential equation known as the Riccati equation. The general solution is y = tan(t + C) .

Applying the initial condition y(0) = 0, we get:

0 = tan(0 + C)

The solution to the IVP is then y = tan(t) .

According to the Existence and Uniqueness Theorem for ordinary differential equations (ODEs) , if the function f and its partial derivative with respect to y are continuous in some region of the (t, y) -plane that includes the initial condition (t₀, y₀) , then there exists a unique solution y(t) to the IVP in some interval about t = t₀ .

In other words, given certain conditions, we are guaranteed to find exactly one solution to the IVP that satisfies both the differential equation and the initial condition .

If a solution exists, it will be a function that is at least once differentiable (since it must satisfy the given ODE ) and, therefore, continuous . The solution will also be differentiable as many times as the order of the ODE .

Small changes in the initial conditions can result in drastically different solutions to an IVP . This is often called “ sensitive dependence on initial conditions ,” a characteristic feature of chaotic systems .

The Existence and Uniqueness Theorem only guarantees a solution in a small interval around the initial point t₀ . This is called a local solution . However, under certain circumstances, a solution might extend to all real numbers, providing a global solution . The nature of the function f and the differential equation itself can limit the interval of the solution.

For higher-order ODEs , you will have more than one initial condition. For an n-th order ODE , you’ll need n initial conditions to find a unique solution.

The solution to an IVP may behave differently as it approaches the boundaries of its validity interval. For example, it might diverge to infinity , converge to a finite value , oscillate , or exhibit other behaviors.

The general solution of an ODE is a family of functions that represent all solutions to the ODE . By applying the initial condition(s), we narrow this family down to one solution that satisfies the IVP .

Applications 

Solving initial value problems (IVPs) is fundamental in many fields, from pure mathematics to physics , engineering , economics , and beyond. Finding a specific solution to a differential equation given initial conditions is essential in modeling and understanding various systems and phenomena. Here are some examples:

IVPs are used extensively in physics . For example, in classical mechanics , the motion of an object under a force is determined by solving an IVP using Newton’s second law ( F=ma , a second-order differential equation). The initial position and velocity (the initial conditions) are used to find a unique solution that describes the object’s motion .

IVPs appear in many engineering problems. For instance, in electrical engineering , they are used to describe the behavior of circuits containing capacitors and inductors . In civil engineering , they are used to model the stress and strain in structures over time.

In biology , IVPs are used to model populations’ growth and decay , the spread of diseases , and various biological processes such as drug dosage and response in pharmacokinetics .

Differential equations model various economic processes , such as capital growth over time. Solving the accompanying IVP gives a specific solution that models a particular scenario, given the initial economic conditions.

IVPs are used to model the change in populations of species , pollution levels in a particular area, and the diffusion of heat in the atmosphere and oceans.

In computer graphics, IVPs are used in physics-based animation to make objects move realistically. They’re also used in machine learning algorithms, like neural differential equations , to optimize parameters.

In control theory , IVPs describe the time evolution of systems. Given an initial state , control inputs are designed to achieve a desired state.

Solve the IVP y’ = 2y, y(0) = 1 .

The given differential equation is separable. Separating variables and integrating, we get:

∫dy/y = ∫2 dt

ln|y| = 2t + C

y = $e^{(2t+C)}$

= $e^C  * e^{(2t)}$

Now, apply the initial condition y(0) = 1 :

1 = $e^C * e^{(2*0)}$

The solution to the IVP is y = e^(2t) .

Solve the IVP y’ = -3y, y(0) = 2 .

The general solution is y = Ce^(-3t) . Apply the initial condition y(0) = 2 to get:

2 = C $e^{(-3*0)}$

2 = C $e^0$

So, C = 2, and the solution to the IVP is y = 2e^(-3t) .

Solve the IVP y’ = y^2, y(1) = 1 .

This is also a separable differential equation. We separate variables and integrate them to get:

∫$dy/y^2$ = ∫dt,

1/y = t + C.

Applying the initial condition y(1) = 1, we find C = -1. So the solution to the IVP is -1/y = t – 1 , or y = -1/(t – 1).

Solve the IVP y” – y = 0, y(0) = 0, y'(0) = 1 .

This is a second-order linear differential equation. The general solution is y = A sin(t) + B cos(t) .

The first initial condition y(0) = 0 gives us:

0 = A 0 + B 1

The second initial condition y'(0) = 1 gives us:

1 = A cos(0) + B*0

The solution to the IVP is y = sin(t) .

Solve the IVP y” + y = 0, y(0) = 1, y'(0) = 0 .

This is also a second-order linear differential equation. The general solution is y = A sin(t) + B cos(t) .

The first initial condition y(0) = 1 gives us:

1 = A 0 + B 1

The second initial condition y'(0) = 0 gives us:

0 = A cos(0) – B*0

The solution to the IVP is y = cos(t) .

Solve the IVP y” = 9y, y(0) = 1, y'(0) = 3.

The differential equation can be rewritten as y” – 9y = 0. The general solution is y = A $ e^{(3t)} + B e^{(-3t)}$ .

1 = A $e^{(30)}$ + B $e^{(-30)}$

So, A + B = 1.

The second initial condition y'(0) = 3 gives us:

3 = 3A $e^{30} $ – 3B $e^{-30}$

= 3A – 3B

So, A – B = 1.

We get A = 1 and B = 0 to solve these two simultaneous equations. So, the solution to the IVP is y = $e^{(3t)}$ .

Solve the IVP y” + 4y = 0, y(0) = 0, y'(0) = 2 .

The differential equation is a standard form of a second-order homogeneous differential equation. The general solution is y = A sin(2t) + B cos(2t) .

The second initial condition y'(0) = 2 gives us:

2 = 2A cos(0) – B*0

The solution to the IVP is y = sin(2t) .

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Initial Value Problem in Calculus | Definition, Formula & Example

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Amy has a master's degree in secondary education and has been teaching math for over 9 years. Amy has worked with students at all levels from those with special needs to those that are gifted.

Simona received her PhD in Applied Mathematics in 2010 and is a college professor teaching undergraduate mathematics courses.

Initial Value when Acceleration Function if Given

If we know the acceleration of a moving particle, as a function of time, we can determine the position function, but we need to know two initial conditions to obtain the unique position function.

Because velocity is an antiderivative of acceleration, we integrate the acceleration function to obtain a velocity function. Integrating the velocity function, we obtain a position function.

With each integration, we have a different constant of integration.

When we integrate the acceleration function, we obtain a velocity function that contains a constant of integration, which is determined by using the velocity at a given point in time.

Integrating the velocity function, we obtain a position function that contains a constant of integration, which is determined uniquely by knowing the position's initial value.

Practice Problem

Find the velocity and position functions of a particle whose acceleration is a(t) = 6t if the particle starts from rest and initially the position is 2 m from the origin.

because the particle starts from rest, v(0)=0 and position is

because, initially s(0)=2.

What is an initial value problem with an example?

An initial value problem is a differential equation with some initial conditions. For example, dy/dx = x with initial conditions y(0)=1.

What's initial value in math?

An initial value problem is a differential equation with some initial conditions that specify the value of the function at some point. The initial value can be used to determine the value of the unknown constant.

How do you solve initial value problems?

There are two steps to solving an initial value problem. The first step is to take the integral of the function. The second step is to use the initial conditions to determine the value of the unknown constant of integration.

Table of Contents

Initial value problem, initial value formula, how to solve initial value problems, initial value problem examples, lesson summary.

Calculus is a type of mathematics that focuses on rates of change, differential calculus, and the summation of many small pieces to make a whole, integral calculus . Initial value problems describe a type of problem in calculus. Initial value problems in calculus concern differential equations with a known initial condition that specifies the value of the function at some point. The purpose of these problems is to find the function that describes the system, which can be done by integrating the differential equation.

When performing an integration with an indefinite integral , or with no initial conditions, there is always an unknown constant. This is known as the constant of integration and is often denoted by {eq}C {/eq}. An example is shown in the diagram.

In the absence of initial conditions, it is not possible to determine the value of this unknown constant. More information is needed to find the value of the constant of integration. Consider the same problem above if initial conditions are provided. Assume the initial conditions are

$$y(0) = 2 $$

This means that at {eq}x=0 {/eq}, the value of the function is 2, {eq}y=2 {/eq}. To determine the constant of integration, these values can be substituted into the function obtained from the integration

$$y = x^2+x+C $$

$$2 = (0)^2+(0)+C $$

Therefore, the constant of integration is 2. The function can then be written as

$$y = x^2+x+2 $$

The initial condition does not have to be when {eq}x=0 {/eq}, it can be given for any value of x. It is called the initial condition because it is the first piece of information known about the function.

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• 0:02 What Is Initial Value?
• 0:41 The Constant of Integration
• 0:57 The Method
• 1:15 Sample Problem
• 2:49 Lesson Summary

Initial value problems are often presented in the same way. This can be thought of as the initial value formula. A derivative function will be defined in terms of two. Initial conditions will then be provided that specifies the value of one variable in relation to another variable. The derivative and the initials are generally given in the form

$$\frac{dy}{dx} = x $$

$$y(0)=0 $$

This information can then be used to solve the initial value problem described.

The next question is how to solve initial value problems. There are two major steps when solving initial value problems. These steps are

• Integrate the differential function to find the function.
• Use the initial conditions to determine the constant of integration.

The integral is often called the antiderivative . This is because the processes reverse each other. In other words, if you take the integral of a function and then the derivative of the result, you are left with the original function. This is illustrated in the diagram.

To explain how to integrate a differential function, consider the example

The first step to integrating this differential function is to move all the x terms to one side. Such that

$$dy = x dx $$

Now, it is possible to integrate both sides of this equation

$$\int dy = \int x dx $$

In more general terms, the integration of a function is given by

$$\int x^n dx = \frac{x^{n+1}}{n+1} +C $$

In the example function above, {eq}n=1 {/eq}. Therefore, the integration gives

$$y = \frac{x^2}{2} + C $$

It is important to remember that each integral will produce a constant of integration. However, both of these have been combined into a single term, {eq}C {/eq}.

After integrating the differential function, the next step is to use initial conditions to determine the constant of integration. This is done by substituting the values given in the initial conditions into the equation. It is then possible to rearrange the equation to find the value of {eq}C {/eq}. For the example above, assume the initial conditions are

$$y(2) = 1 $$

Substituting these initial conditions into the function gives

$$1 = \frac{(2)^2}{2} + C $$

Therefore, {eq}C = -1 {/eq}. So, the solution to the initial value problem is

$$y = \frac{x^2}{2}-1 $$

In order to get a better understanding, some initial value problem examples may be useful. Here are three initial value example problems.

First, consider the initial value problem described by

$$\frac{dy}{dx} = 6x-2 $$

$$y(2)=0 $$

As mentioned above, the first step is to find the integral of the function. Rearranging the differential equation gives

$$dy = (6x-2)dx $$

Then taking the integral of both sides

$$\int dy = \int (6x-2) dx $$

So, the function is given by

$$y = 3x^2 -2x+C $$

The next step is to use the initial conditions to determine the value of the constant of integration. This is done by substituting the initial values given

$$0 = 3(2)^2-2(2)+C $$

Therefore, the constant of integration is

And the function can be written as

$$y=3x^2-2x-8 $$

Consider the initial value problem described by

$$\frac{dy}{dx} = 3x^2+4x-1 $$

$$y(1)=9 $$

Integrating this function gives

$$y = x^3+\frac{4x^2}{2}-x+C $$

The next step is to substitute the values from the initial conditions, which gives

$$9 = (1)^3+\frac{4(1)^2}{2}-(1)+C $$

This gives the constant of integration

{eq}C=7 {/eq}. Therefore, the solution is

$$y = x^3+\frac{4x^2}{2}-x+7 $$

One final example that does not follow the general formula for integration described above. Consider the initial value problem

$$\frac{dy}{dx}=cos(x) $$

$$y(\pi)=1 $$

The integral of a cosine function is a sine function, therefore

$$y = sin(x)+C $$

Now substituting the initial conditions gives

$$1 = sin(\pi)+C $$

This gives a constant of integration, {eq}C = 1 {/eq}. So, the function is

$$y = sin(x)+1 $$

In calculus, initial value problems involve taking the integral of a derivative function. When an integration is performed, an unknown constant of integration is left. For initial value problems , some initial conditions specify the value of the function at some point. They are generally given in the form {eq}y(0)=1 {/eq}. This means that when {eq}x=0 {/eq}, {eq}y=1 {/eq}. These values can be substituted into a function to determine the value of the constant. Despite the name, the initial conditions do not need to be given at {eq}x=0 {/eq}, they can be any point. The first step for solving an initial value problem is to integrate the function. The general formula for integration is

However, when considering problems that involve trigonometric there are rules to remember. For example, {eq}\int cos(x) = sin(x) +C {/eq}. After the function has been integrated, the initial conditions can be substituted to determine the value of the unknown constant.

Video Transcript

What is initial value.

Initial value in calculus is a type of problem involving the use of an initial condition. This type of problem produces an unknown constant that requires the use of an initial condition or known point to solve. When you are given an initial condition, it will look like this:

This tells you that when x = 0, your y = 2. The initial condition does not have to be when x = 0. It can be any point. It's called an initial condition because it is the first bit of information you know about the newly integrated function. You need this information to fully integrate functions. Let's see why.

The Constant of Integration

When you calculate the indefinite integral, you end up with something called the constant of integration. It looks like this when you write it out.

Because you have this unknown constant, you need a known point to plug into your equation to figure it out. This known point is your initial condition.

The method to solving initial value problems requires just a couple of steps:

• Integrate the derivative to find the function.
• Use your initial condition to find your constant of integration to get your full answer.

Let's see how these two steps play out with a sample problem.

Sample Problem

The integral is also called the antiderivative because if you take the derivative of the answer, you would get the function you just integrated. With initial value problems, you're often given the derivative of the function you are trying to find along with an initial condition.

Our first step tells us to integrate, or take the antiderivative of the derivative. To do that, we need to move the dx over to the other side. So, let's integrate both sides.

We have combined the constant of integration from both sides into just the one because they are both constants and are therefore, like terms. They'll be combined later. Our answer isn't complete without knowing this constant. So, let's go onto step two.

Step two tells us to use the initial condition to help us find the constant of integration. Our initial condition tells us that when x = 0, our y = 2. Let's plug that information into our integration result and solve for the constant.

Our constant of integration is a 2. Now we can plug this information into our integration result to get our full answer.

Wow! We are done. That wasn't so bad. Just two steps to solve this type of problem!

Let's review. Initial value problems involve the use of an initial condition to help you solve integration problems where you have a constant of integration. There are two steps involved in solving such problems. The two steps are to integrate and then to use the initial condition to find the constant of integration to get the full answer.

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Laplace Transform

4.4 Solving Initial Value Problems

Having explored the Laplace Transform, its inverse, and its properties, we are now equipped to solve initial value problems (IVP) for linear differential equations. Our focus will be on second-order linear differential equations with constant coefficients.

Method of Laplace Transform for IVP

General Approach:

1. Apply the Laplace Transform to each term of the differential equation. Use the properties of the Laplace Transform listed in Tables 4.1 and 4.2 to obtain an equation in terms of [asciimath]Y(s)[/asciimath] . The Laplace Transform of the derivatives are

  [asciimath]\mathcal{L}{f'(t)} = sF(s) - f(0)[/asciimath]

  [asciimath]\mathcal{L}{f''(t)\} = s^2F(s) - s f(0) - f'(0)[/asciimath]

2. The transforms of derivatives involve initial conditions at [asciimath]t=0[/asciimath] . Apply the initial conditions.

3. Simplify the transformed equation to isolate  [asciimath]Y(s)[/asciimath] .

4. If needed, use partial fraction decomposition to break down  [asciimath]Y(s)[/asciimath] into simpler components.

5. Determine the inverse Laplace Transform using the tables and linearity property to find [asciimath]y(t)[/asciimath] .

Shortcut Approach:

1. Find the characteristic polynomial of the differential equation [asciimath]p(s)=as^2+bs+c[/asciimath] .

2. Substitute [asciimath]p(s)[/asciimath] , [asciimath]F(s)=\mathcal{L}{f(t)}[/asciimath] , and the initial conditions into the equation

  [asciimath]Y(s)=(F(s)+a(y'(0)+sy(0))+b y(0) )/(p(s))[/asciimath]   (4.4.1)

3. If needed, use partial fraction decomposition to break down [asciimath]Y(s)[/asciimath] into simpler components.

4. Determine the inverse Laplace transform of [asciimath]Y(s)[/asciimath] using the tables and linearity property to find [asciimath]y(t)[/asciimath] .

Solve the initial value problem.

  [asciimath]y''-5y'+6y=4e^(-2t)\ ;[/asciimath]      [asciimath]y(0)=-1, \ y'(0)=2[/asciimath]

Using the General Approach

1. Take the Laplace Transform of both sides of the equation

  [asciimath]\mathcal{L}^-1{ y''}-5\mathcal{L}^-1{ y'}+6\mathcal{L}^-1{y}=4\mathcal{L}^-1{ e^(-2t)}[/asciimath]

Letting [asciimath]Y(s)=\mathcal{L}^-1{y}[/asciimath] , we get

  [asciimath]s^2Y(s)-sy(0)-y'(0)-5(sY(s)-y(0))+6Y(s)=4(1/(s+2))[/asciimath]

2. Plugging in the initial conditions gives

  [asciimath]s^2Y(s)+s-2-5(sY(s)+1)+6Y(s)=4(1/(s+2))[/asciimath]

3. Collecting like terms and isolating [asciimath]Y(s)[/asciimath] , we get

  [asciimath](s^2-5s+6)Y(s)=4/(s+2)-s+7[/asciimath]

  [asciimath]Y(s)[/asciimath]   [asciimath]=(4//(s+2)-s+7)/(s^2-5s+6)[/asciimath]

Multiplying both the denominator and numerator by [asciimath](s+2)[/asciimath] and factoring the denominator yields

  [asciimath]Y(s)=(-s^2+5s+18)/((s+2)(s-3)(s-2))[/asciimath]

4. Using partial fraction expansion, we get

  [asciimath]Y(s)=1/5 (1/(s+2))+24/5 (1/(s-3))-6 (1/(s-2))[/asciimath]

5. From Table 4.1 , we see that

  [asciimath]1/(s-a)[/asciimath]      [asciimath]harr \ \e^(at)[/asciimath]

Taking the inverse, we obtain the solution of the equation

  [asciimath]y(t)=\mathcal{L}^-1{Y(s)}[/asciimath]   [asciimath]=1/5 \ e^(-2t) +24/5 e^(3t)-6 e^(2t)[/asciimath]

  [asciimath]y''+4y=3sin(t) \ ;[/asciimath]      [asciimath]y(0)=1, \ y'(0)=-1[/asciimath]

Using the Shortcut Approach

1. The characteristic polynomial is

  [asciimath]p(s)=s^2+4[/asciimath]

  [asciimath]F(s)=\mathcal{L}^-1{3sin(t)}[/asciimath]   [asciimath]=3/(s^2+1)[/asciimath]

2. Substituting them together with the initial values into Equation 4.4.1 , we obtain

  [asciimath]Y(s)=(3//(s^2+1)+(-1+s(1)))/(s^2+4)[/asciimath] [asciimath]=(3//(s^2+1)+s-1)/(s^2+4)[/asciimath]

Multiplying both the denominator and numerator by [asciimath](s^2+1)[/asciimath]  yields

  [asciimath]Y(s)=(s^3-s^2+s+2)/((s^2+1)(s^2+4))[/asciimath]

3. Using partial fraction expansion, we get

  [asciimath]Y(s)=1/(s^2+1)+(s-2)/(s^2+4)[/asciimath]

[asciimath]\ =1/(s^2+1)+s/(s^2+4)- 2/(s^2+4)[/asciimath]

4. From Table 4.1 ,

  [asciimath]sin(bt)\ harr\ b/(s^2+b^2)[/asciimath]      and    [asciimath]cos(bt)\ harr\ s/(s^2+b^2)[/asciimath]

  [asciimath]y(t)=\mathcal{L}^-1{Y(s)}[/asciimath]   [asciimath]=sin(t)+cos(2t)-sin(2t)[/asciimath]

Section 4.4 Exercises

  [asciimath]y'' +3 y' -10 y = 0, \ quad y(0) = -1, \ quad y'(0) = 2[/asciimath]

[asciimath]y(t)=-3/7 e^(2t)-4/7 e^(-5t)[/asciimath]

  [asciimath]y'' +6 y' + 13 y = 0, \ quad y(0) = 2, \ quad y'(0) = 0[/asciimath]

[asciimath]y(t)=e^(-3t)(2cos(2t)+3sin(2t))[/asciimath]

  [asciimath]y'' - 8 y' +16 y = 0, \ quad y(0) = 1, \ quad y'(0) = -1[/asciimath]

  [asciimath]y(t)=e^(4t)(1-5t)[/asciimath]

Differential Equations Copyright © 2024 by Amir Tavangar is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Euler's Method Explained with Examples

What is Euler’s Method?

The Euler’s method is a first-order numerical procedure for solving ordinary differential equations (ODE) with a given initial value.

The General Initial Value Problem

Methodology

Euler’s method uses the simple formula,

to construct the tangent at the point x and obtain the value of y(x+h) , whose slope is,

In Euler’s method, you can approximate the curve of the solution by the tangent in each interval (that is, by a sequence of short line segments), at steps of h .

In general , if you use small step size, the accuracy of approximation increases.

General Formula

Functional value at any point b , given by y(b)

• n = number of steps
• h = interval width (size of each step)

Find y(1) , given

Solving analytically, the solution is y = e x and y(1) = 2.71828 . (Note: This analytic solution is just for comparing the accuracy.)

Using Euler’s method, considering h = 0.2 , 0.1 , 0.01 , you can see the results in the diagram below.

When h = 0.2 , y(1) = 2.48832 (error = 8.46 %)

When h = 0.1 , y(1) = 2.59374 (error = 4.58 %)

When h = 0.01 , y(1) = 2.70481 (error = 0.50 %)

You can notice, how accuracy improves when steps are small.

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Number Line

The Calculus Calculator is a powerful online tool designed to assist users in solving various calculus problems efficiently. Here's how to make the most of its capabilities:

• Begin by entering your mathematical expression into the above input field, or scanning it with your camera.
• Choose the specific calculus operation you want to perform, such as differentiation, integration, or finding limits.
• Once you've entered the function and selected the operation, click the 'Go' button to generate the result.
• The calculator will instantly provide the solution to your calculus problem, saving you time and effort.
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• \lim _{x\to \:0}(\frac{\sin (x)}{x})
• \int e^x\cos (x)dx
• \int \cos^3(x)\sin (x)dx
• \int_{0}^{\pi}\sin(x)dx
• \frac{d}{dx}(\frac{3x+9}{2-x})
• \frac{d^2}{dx^2}(\frac{3x+9}{2-x})
• implicit\:derivative\:\frac{dy}{dx},\:(x-y)^2=x+y-1
• \sum_{n=0}^{\infty}\frac{3}{2^n}
• tangent\:of\:f(x)=\frac{1}{x^2},\:(-1,\:1)

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What Is Solution Selling? A Complete Guide

Solution selling is one of the most effective approaches to selling regardless of business size.

Erin Hueffner

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Salespeople who push a particular product or service before they understand the customer give the whole industry a bad reputation. Our  State of Sales  research shows that 87% of business buyers expect sales reps to act as trusted advisors. That’s why solution selling is such an important technique for you to master. You show your prospects you fully understand the challenges of their business and that you’re here to help solve them.

Let’s look at how this proven sales approach can help you hit — and surpass — sales targets.

What you’ll learn:

What is solution selling, how does solution selling differ from product selling, why is solution selling important, when should you use solution selling, what are the steps of solution selling, pros and cons of solution selling, level up your game with ai conversation insights.

Sell smarter using Einstein Conversation Insights — with customer signals and next-step guidance to help you close.

Solution selling is a sales technique that focuses on your customers’ needs and pain points and provides recommendations to solve them.

In the years since this method was introduced to the world of sales, solution selling has become a customer-centric alternative to the previously standard approach of “box pushing.” Where box pushing focuses on selling product features and specs in, more or less, the same way to all prospects, solution selling brought a whole new level of customization and consultation that focuses on each customer’s specific needs.

Solution selling focuses on the customer and the problems they’re experiencing. Like box pushing, product selling focuses on the product and selling as many of them as possible — regardless of whether or not the customer needs the product. This shift in perspective means that solution selling is a more consultative and relationship-focused process than product selling.

For example, product selling promotes a specific product or service by emphasizing its features and benefits and working through a script that pushes the product. Customers may or may not have a clear problem or need that the product solves; the focus is solely on selling. There’s also very little follow-up or consultation involved in the process.

Solution selling, on the other hand, uses active listening to uncover what a customer’s problems are, then offers recommendations that address those problems. It’s very consultative, and can lead to a long-term relationship and repeat business.

At its heart, solution selling is all about the ROI that your prospect can get out of your recommendations, not pushing features on them in hopes of making a sale. This builds trust-based relationships and meets the prospect’s need to work with someone who puts their best interest at heart.

Rather than focusing on your product’s features and benefits, solution selling is centered around your prospects’ needs:

• What are their goals and pain points?
• What problems and challenges are they facing?
• What is the outcome that can solve their needs?

Solution selling means being both  empathetic and practical . The seller should go beyond the surface-level handshake and really understand the buyer’s industry, challenges, and goals. When you walk in your customer’s shoes and understand their pain points from the inside out, you are much more qualified to tailor the right solution to them.

There’s a cliche that sales is all about building rapport based on small talk. Solution selling goes deeper. Rapport is based on knowing your customer. Maybe they’re about to have a merger, or they’re experiencing challenges with the supply chain. The solution seller’s role is to provide insight that helps customers see a vision of a better future.

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Solution selling can be an effective way to approach any sales scenario. However, it’s most effective when you’re working with prospects who need a solution to a unique or “niche” problem. This often occurs in industries that feature the following characteristics:

• High-value sales and long sales cycles focused on software systems, industrial or manufacturing equipment, or IT consulting
• An inventory marked by evolving technology such as telecommunications and healthcare
• Significant regulatory compliance requirements, such as those in the pharmaceutical and finance industries

It’s called “solution” selling and not “product” selling for a reason: you’re selling a solution, a recommendation, or an outcome. To talk about an outcome, you need to understand the challenges and needs of where your prospect is now.

Here’s a series of steps you can follow when perfecting your own solution sales techniques:

1. Understand your customer

Effective sales conversations only happen when you understand your customer. So, start there!

First, explore your company’s  buyer personas  to familiarize yourself with the type of organization, person, and pain points you’ll be dealing with most often. This information is theoretical, and you’ll need to make adjustments when you interact with a real client, but it’s a smart place to start.

Next, read up on the industry and research particular customers you plan to meet. Your customer relationship management product may help you do this using  artificial intelligence (AI).

Finally, use your discovery call with a potential client to ask leading questions and learn more about them. Start with the following and see where the conversation takes you:

• What is your biggest priority this year?
• What obstacles stand in your way?
• What solutions have you already tried?
• What do you stand to lose if you can’t fix this problem?
• What do you stand to gain if you can?

By allowing your understanding of the customer to develop 1:1, you’ll capture a clear and detailed picture of the customer’s circumstances and needs. You’ll then be able to assess how your solution will fit into the bigger picture.

2. Understand your products

Solution selling may not be about selling product features, but it’s still important to understand them. After all, your product’s features are the reason you’re able to solve your customer problems. Once you have an overview of your customer’s basic needs, you’ll have the context you need to be able to map your products’ features to those challenges to identify potential solutions.

Here are a few things you can do to quickly build up your understanding of your product’s features:

• Attend a live or pre-recorded demo of your product. A live or recorded product demo gives you a good sense of how customers are introduced to your product. It will also walk you through the most straightforward use cases of the product for you to learn.
• Listen to recorded sales calls or customer calls. Many organizations use call recording software to capture real-life customer feedback. Incorporate “listening tours” with customers into your routine so you can stay up-to-date with how they perceive your product.
• Take notes. Build a habit of taking notes and summarizing what you learn from prospects in a notebook or digital app. Reflect on these notes weekly and curate your own body of knowledge about your product’s features.

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3. show clients exactly what they’re missing.

After completing the first two steps, you’ll be well on your way to a solution selling process that’s tailored to your customer’s needs, and not just a generic list of features on a sales sheet. Why? Because you’ll have a deep understanding of the problems they’re experiencing, and you’ll understand the priority of those problems. You can then map each problem to a solution or feature available in your product, and paint a picture of how this problem can be solved holistically.

For example, let’s say you’re selling a cloud-based business technology solution to a  small manufacturing business.  Their sales reps keep losing deals because of supply chain forecasting issues. The real problem? They don’t have real-time inventory information to react and adjust proactively.

Here’s what your product offers that can serve as a solution: Real-time inventory information for every stakeholder and a single source of truth will ensure they’re always on same page. With this, they can efficiently identify and get ahead of any challenges related to the supply chain. As you share these details, be sure to show them how they can succeed with your partnership, don’t tell them how your product works.

4. Close the deal and maintain the relationship for the future

The relationship-focused approach of solution selling continues throughout the  sales funnel  all the way up to closing the deal and beyond. In fact, our State of Sales Report highlights that the ongoing relationship with a customer is a growing area of focus for sales teams:  80% of sales reps  say maintaining customer relationships after the close is increasingly important, and 92% say they are at least partially evaluated on post-sale metrics like customer value and customer retention

Every sales strategy has benefits and drawbacks, and solution selling is no exception. If you’re thinking about this approach, here are some pros and cons to consider.

Pros of solution selling

• Focuses on the customer’s long-term success.  Solution selling focuses on the customer’s long-term vision for their business and emphasizes eliminating challenges that are limiting their growth. This approach builds up a partnership based on trust and mutual benefit, rather than a temporary, transactional relationship. By working with you, your customer gains access to more efficiency in their business, and you capture a potentially recurring source of revenue — it’s the definition of win-win.
• Customized to each customer.  Solution selling grows out of a 1:1 relationship with a customer, resulting in a well-designed, well-implemented recommendation that’s unique and customized to each customer. This should lead to a happier customer who is more likely to return for your input and recommendations again and again.

Cons of solution selling

• Time commitment. Solution selling works for almost every sales scenario. But because you must prioritize long-term value and ROI over short-term gains, you will need to invest significant time and energy in building recommendations and a long-term relationship with a prospect.
• High-touch. Solution selling requires the salesperson to be intimately familiar with the customer and engage in in-depth conversations about their business needs, ultimately requiring more per-customer research and time in conversation than other kinds of selling.
• Longer sales cycles. Partly because of the increased frequency of meetings and partly because of the customization of the solution itself, solution selling can require a longer sales cycle or cadence than other kinds of selling.

Build trusted relationships with solution selling

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8.3: Solution of Initial Value Problems

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• Page ID 30766

• William F. Trench
• Trinity University

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Laplace Transforms of Derivatives

In the rest of this chapter we’ll use the Laplace transform to solve initial value problems for constant coefficient second order equations. To do this, we must know how the Laplace transform of \(f'\) is related to the Laplace transform of \(f\). The next theorem answers this question.

Theorem 8.3.1

Suppose \(f\) is continuous on \([0,\infty)\) and of exponential order \(s_0\), and \(f'\) is piecewise continuous on \([0,\infty).\) Then \(f\) and \(f'\) have Laplace transforms for \(s > s_0,\) and

\[\label{eq:8.3.1} {\mathscr L}(f')=s {\mathscr L}(f)-f(0).\]

We know from Theorem 8.1.6 that \({\mathscr L}(f)\) is defined for \(s>s_0\). We first consider the case where \(f'\) is continuous on \([0,\infty)\). Integration by parts yields

\[ \begin{align} \int^T_0 e^{-st}f'(t)\,dt &= e^{-st}f(t)\Big|^T_0+s \int^T_0e^{-st}f(t)\,dt \nonumber \\[4pt] &= e^{-sT}f(T)-f(0)+s\int^T_0 e^{-st}f(t)\,dt \label{eq:8.3.2} \end{align} \]

for any \(T>0\). Since \(f\) is of exponential order \(s_0\), \(\displaystyle \lim_{T\to \infty}e^{-sT}f(T)=0\) and the integral in on the right side of Equation \ref{eq:8.3.2} converges as \(T\to\infty\) if \(s> s_0\). Therefore

\[\begin{aligned} \int^\infty_0 e^{-st}f'(t)\,dt&=-f(0)+s\int^\infty_0 e^{-st}f(t)\,dt\\ &=-f(0)+s{\mathscr L}(f),\end{aligned}\nonumber \]

which proves Equation \ref{eq:8.3.1}.

Suppose \(T>0\) and \(f'\) is only piecewise continuous on \([0,T]\), with discontinuities at \(t_1 < t_2 <\cdots < t_{n-1}\). For convenience, let \(t_0=0\) and \(t_n=T\). Integrating by parts yields

\[\begin{aligned} \int^{t_i}_{t_{i-1}}e^{-st}f'(t)\,dt &= e^{-st}f(t)\Big|^{t_i}_{t_{i-1}}+s\int^{t_i}_{t_{i-1}}e^{-st}f(t)\,dt\\ &= e^{-st_i} f(t_i)- e^{-st_{i-1}}f(t_{i-1})+s\int^{t_i}_{t_{i-1}}e^{-st}f(t)\,dt.\end{aligned}\nonumber\]

Summing both sides of this equation from \(i=1\) to \(n\) and noting that

\[\left(e^{-st_1}f(t_1)-e^{-st_0}f(t_0)\right)+\left(e^{-st_2} f(t_2)-e^{-st_1}f(t_1)\right) +\cdots+\left(e^{-st_N}f(t_N)-e^{-st_{N-1}}f(t_{N-1})\right) \nonumber\]

\[=e^{-st_N}f(t_N)-e^{-st_0}f(t_0)=e^{-sT}f(T)-f(0) \nonumber\]

yields Equation \ref{eq:8.3.2}, so Equation \ref{eq:8.3.1} follows as before.

Example 8.3.1

In Example 8.1.4 we saw that

\[{\mathscr L}(\cos\omega t)={s\over s^2+\omega^2}. \nonumber\]

Applying Equation \ref{eq:8.3.1} with \(f(t)=\cos\omega t\) shows that

\[{\mathscr L}(-\omega\sin\omega t)=s {s\over s^2+\omega^2}-1=- {\omega^2\over s^2+\omega^2}.\nonumber\]

\[{\mathscr L}(\sin\omega t)={\omega\over s^2+\omega^2},\nonumber\]

which agrees with the corresponding result obtained in 8.1.4.

In Section 2.1 we showed that the solution of the initial value problem

\[\label{eq:8.3.3} y'=ay, \quad y(0)=y_0,\]

is \(y=y_0e^{at}\). We’ll now obtain this result by using the Laplace transform.

Let \(Y(s)={\mathscr L}(y)\) be the Laplace transform of the unknown solution of Equation \ref{eq:8.3.3}. Taking Laplace transforms of both sides of Equation \ref{eq:8.3.3} yields

\[{\mathscr L}(y')={\mathscr L}(ay),\nonumber\]

which, by Theorem 8.3.1 , can be rewritten as

\[s{\mathscr L}(y)-y(0)=a{\mathscr L}(y),\nonumber\]

\[sY(s)-y_0=aY(s).\nonumber\]

Solving for \(Y(s)\) yields

\[Y(s)={y_0\over s-a},\nonumber\]

\[y={\mathscr L}^{-1}(Y(s))={\mathscr L}^{-1}\left({y_0\over s-a}\right)=y_0{\mathscr L}^{-1}\left({1\over s-a}\right)=y_0e^{at},\nonumber\]

which agrees with the known result.

We need the next theorem to solve second order differential equations using the Laplace transform.

Theorem 8.3.2

Suppose \(f\) and \(f'\) are continuous on \([0,\infty)\) and of exponential order \(s_0,\) and that \(f''\) is piecewise continuous on \([0,\infty).\) Then \(f\), \(f'\), and \(f''\) have Laplace transforms for \(s > s_0\),

\[\label{eq:8.3.4} {\mathscr L}(f')=s {\mathscr L}(f)-f(0),\]

\[\label{eq:8.3.5} {\mathscr L}(f'')=s^2{\mathscr L}(f)-f'(0)-sf(0).\]

Theorem 8.3.1 implies that \({\mathscr L}(f')\) exists and satisfies Equation \ref{eq:8.3.4} for \(s>s_0\). To prove that \({\mathscr L}(f'')\) exists and satisfies Equation \ref{eq:8.3.5} for \(s>s_0\), we first apply Theorem 8.3.1 to \(g=f'\). Since \(g\) satisfies the hypotheses of Theorem 8.3.1 , we conclude that \({\mathscr L}(g')\) is defined and satisfies

\[{\mathscr L}(g')=s{\mathscr L}(g)-g(0)\nonumber\]

for \(s>s_0\). However, since \(g'=f''\), this can be rewritten as

\[{\mathscr L}(f'')=s{\mathscr L}(f')-f'(0).\nonumber\]

Substituting Equation \ref{eq:8.3.4} into this yields Equation \ref{eq:8.3.5}.

Solving Second Order Equations with the Laplace Transform

We’ll now use the Laplace transform to solve initial value problems for second order equations.

Example 8.3.2

Use the Laplace transform to solve the initial value problem

\[\label{eq:8.3.6} y''-6y'+5y=3e^{2t},\quad y(0)=2, \quad y'(0)=3.\]

Taking Laplace transforms of both sides of the differential equation in Equation \ref{eq:8.3.6} yields

\[{\mathscr L}(y''-6y'+5y)={\mathscr L}\left(3e^{2t}\right)={3\over s-2},\nonumber\]

which we rewrite as

\[\label{eq:8.3.7} {\mathscr L}(y'')-6{\mathscr L}(y')+5{\mathscr L}(y)={3\over s-2}.\]

Now denote \({\mathscr L}(y)=Y(s)\). Theorem 8.3.2 and the initial conditions in Equation \ref{eq:8.3.6} imply that

\[{\mathscr L}(y')=sY(s)-y(0)=sY(s)-2\nonumber\]

\[{\mathscr L}(y'')=s^2Y(s)-y'(0)-sy(0)=s^2Y(s)-3-2s.\nonumber\]

Substituting from the last two equations into Equation \ref{eq:8.3.7} yields

\[\left(s^2Y(s)-3-2s\right)-6\left(sY(s)-2\right)+5Y(s)={3\over s-2}.\nonumber\]

\[\label{eq:8.3.8} (s^2-6s+5)Y(s)={3\over s-2}+(3+2s)+6(-2),\]

\[(s-5)(s-1)Y(s)={3+(s-2)(2s-9)\over s-2},\nonumber\]

\[Y(s)={3+(s-2)(2s-9)\over(s-2)(s-5)(s-1)}.\nonumber\]

Heaviside’s method yields the partial fraction expansion

\[Y(s)=-{1\over s-2}+{1\over2}{1\over s-5}+{5\over2}{1\over s-1},\nonumber\]

and taking the inverse transform of this yields

\[y=-e^{2t}+{1\over2}e^{5t}+{5\over2}e^t \nonumber\]

as the solution of Equation \ref{eq:8.3.6}.

It isn’t necessary to write all the steps that we used to obtain Equation \ref{eq:8.3.8}. To see how to avoid this, let’s apply the method of Example 8.3.2 to the general initial value problem

\[\label{eq:8.3.9} ay''+by'+cy=f(t), \quad y(0)=k_0,\quad y'(0)=k_1.\]

Taking Laplace transforms of both sides of the differential equation in Equation \ref{eq:8.3.9} yields

\[\label{eq:8.3.10} a{\mathscr L}(y'')+b{\mathscr L}(y')+c{\mathscr L}(y)=F(s).\]

Now let \(Y(s)={\mathscr L}(y)\). Theorem 8.3.2 and the initial conditions in Equation \ref{eq:8.3.9} imply that

\[{\mathscr L}(y')=sY(s)-k_0\quad \text{and} \quad {\mathscr L}(y'')=s^2Y(s)-k_1-k_0s.\nonumber\]

Substituting these into Equation \ref{eq:8.3.10} yields

\[\label{eq:8.3.11} a\left(s^2Y(s)-k_1-k_0s\right)+b\left(sY(s)-k_0\right)+cY(s)=F(s).\]

The coefficient of \(Y(s)\) on the left is the characteristic polynomial

\[p(s)=as^2+bs+c\nonumber\]

of the complementary equation for Equation \ref{eq:8.3.9}. Using this and moving the terms involving \(k_0\) and \(k_1\) to the right side of Equation \ref{eq:8.3.11} yields

\[\label{eq:8.3.12} p(s)Y(s)=F(s)+a(k_1+k_0s)+bk_0.\]

This equation corresponds to Equation \ref{eq:8.3.8} of Example 8.3.2 . Having established the form of this equation in the general case, it is preferable to go directly from the initial value problem to this equation. You may find it easier to remember Equation \ref{eq:8.3.12} rewritten as

\[\label{eq:8.3.13} p(s)Y(s)=F(s)+a\left(y'(0)+sy(0)\right)+by(0).\]

Example 8.3.3

\[\label{eq:8.3.14} 2y''+3y'+y=8e^{-2t}, \quad y(0)=-4,\; y'(0)=2.\]

The characteristic polynomial is

\[p(s)=2s^2+3s+1=(2s+1)(s+1)\nonumber\]

\[F(s)={\mathscr L}(8e^{-2t})={8\over s+2},\nonumber\]

so Equation \ref{eq:8.3.13} becomes

\[(2s+1)(s+1)Y(s)={8\over s+2}+2(2-4s)+3(-4).\nonumber\]

\[Y(s)={4\left(1-(s+2)(s+1)\right)\over (s+1/2)(s+1)(s+2)}.\nonumber\]

\[Y(s)={4\over3}{1\over s+1/2}-{8\over s+1}+{8\over3}{1\over s+2},\nonumber\]

so the solution of Equation \ref{eq:8.3.14} is

\[y={\mathscr L}^{-1}(Y(s))={4\over3}e^{-t/2}-8e^{-t}+{8\over3}e^{-2t}\nonumber\]

(Figure 8.3.1 ).

Example 8.3.4

Solve the initial value problem

\[\label{eq:8.3.15} y''+2y'+2y=1, \quad y(0)=-3,\; y'(0)=1.\]

\[p(s)=s^2+2s+2=(s+1)^2+1\nonumber\]

\[F(s)={\mathscr L}(1)={1\over s},\nonumber\]

\[\left[(s+1)^2+1\right] Y(s)={1\over s}+1\cdot(1-3s)+2(-3).\nonumber\]

\[Y(s)={1-s(5+3s)\over s\left[(s+1)^2+1\right]}.\nonumber\]

In Example 8.2.8 we found the inverse transform of this function to be

\[y={1\over2}-{7\over2}e^{-t}\cos t-{5\over2}e^{-t}\sin t \nonumber\]

(Figure 8.3.2 ), which is therefore the solution of Equation \ref{eq:8.3.15}.

In our examples we applied Theorems 8.3.1 and 8.3.2 without verifying that the unknown function \(y\) satisfies their hypotheses. This is characteristic of the formal manipulative way in which the Laplace transform is used to solve differential equations. Any doubts about the validity of the method for solving a given equation can be resolved by verifying that the resulting function \(y\) is the solution of the given problem.

How To Start A Business In 11 Steps (2024 Guide)

Updated: Apr 7, 2024, 1:44pm

Table of Contents

Before you begin: get in the right mindset, 1. determine your business concept, 2. research your competitors and market, 3. create your business plan, 4. choose your business structure, 5. register your business and get licenses, 6. get your finances in order, 7. fund your business, 8. apply for business insurance, 9. get the right business tools, 10. market your business, 11. scale your business, what are the best states to start a business, bottom line, frequently asked questions (faqs).

Starting a business is one of the most exciting and rewarding experiences you can have. But where do you begin? There are several ways to approach creating a business, along with many important considerations. To help take the guesswork out of the process and improve your chances of success, follow our comprehensive guide on how to start a business. We’ll walk you through each step of the process, from defining your business idea to registering, launching and growing your business.

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The public often hears about overnight successes because they make for a great headline. However, it’s rarely that simple—they don’t see the years of dreaming, building and positioning before a big public launch. For this reason, remember to focus on your business journey and don’t measure your success against someone else’s.

Consistency Is Key

New business owners tend to feed off their motivation initially but get frustrated when that motivation wanes. This is why it’s essential to create habits and follow routines that power you through when motivation goes away.

Take the Next Step

Some business owners dive in headfirst without looking and make things up as they go along. Then, there are business owners who stay stuck in analysis paralysis and never start. Perhaps you’re a mixture of the two—and that’s right where you need to be. The best way to accomplish any business or personal goal is to write out every possible step it takes to achieve the goal. Then, order those steps by what needs to happen first. Some steps may take minutes while others take a long time. The point is to always take the next step.

Most business advice tells you to monetize what you love, but it misses two other very important elements: it needs to be profitable and something you’re good at. For example, you may love music, but how viable is your business idea if you’re not a great singer or songwriter? Maybe you love making soap and want to open a soap shop in your small town that already has three close by—it won’t be easy to corner the market when you’re creating the same product as other nearby stores.

If you don’t have a firm idea of what your business will entail, ask yourself the following questions:

• What do you love to do?
• What do you hate to do?
• Can you think of something that would make those things easier?
• What are you good at?
• What do others come to you for advice about?
• If you were given ten minutes to give a five-minute speech on any topic, what would it be?
• What’s something you’ve always wanted to do, but lacked resources for?

These questions can lead you to an idea for your business. If you already have an idea, they might help you expand it. Once you have your idea, measure it against whether you’re good at it and if it’s profitable.

Your business idea also doesn’t have to be the next Scrub Daddy or Squatty Potty. Instead, you can take an existing product and improve upon it. You can also sell a digital product so there’s little overhead.

What Kind of Business Should You Start?

Before you choose the type of business to start, there are some key things to consider:

• What type of funding do you have?
• How much time do you have to invest in your business?
• Do you prefer to work from home or at an office or workshop?
• What interests and passions do you have?
• Can you sell information (such as a course), rather than a product?
• What skills or expertise do you have?
• How fast do you need to scale your business?
• What kind of support do you have to start your business?
• Are you partnering with someone else?
• Does the franchise model make more sense to you?

Consider Popular Business Ideas

Not sure what business to start? Consider one of these popular business ideas:

• Start a Franchise
• Start a Blog
• Start an Online Store
• Start a Dropshipping Business
• Start a Cleaning Business
• Start a Bookkeeping Business
• Start a Clothing Business
• Start a Landscaping Business
• Start a Consulting Business
• Start a Photography Business
• Start a Vending Machine Business

Most entrepreneurs spend more time on their products than they do getting to know the competition. If you ever apply for outside funding, the potential lender or partner wants to know: what sets you (or your business idea) apart? If market analysis indicates your product or service is saturated in your area, see if you can think of a different approach. Take housekeeping, for example—rather than general cleaning services, you might specialize in homes with pets or focus on garage cleanups.

Primary Research

The first stage of any competition study is primary research, which entails obtaining data directly from potential customers rather than basing your conclusions on past data. You can use questionnaires, surveys and interviews to learn what consumers want. Surveying friends and family isn’t recommended unless they’re your target market. People who say they’d buy something and people who do are very different. The last thing you want is to take so much stock in what they say, create the product and flop when you try to sell it because all of the people who said they’d buy it don’t because the product isn’t something they’d buy.

Secondary Research

Utilize existing sources of information, such as census data, to gather information when you do secondary research. The current data may be studied, compiled and analyzed in various ways that are appropriate for your needs but it may not be as detailed as primary research.

Conduct a SWOT Analysis

SWOT stands for strengths, weaknesses, opportunities and threats. Conducting a SWOT analysis allows you to look at the facts about how your product or idea might perform if taken to market, and it can also help you make decisions about the direction of your idea. Your business idea might have some weaknesses that you hadn’t considered or there may be some opportunities to improve on a competitor’s product.

Asking pertinent questions during a SWOT analysis can help you identify and address weaknesses before they tank your new business.

A business plan is a dynamic document that serves as a roadmap for establishing a new business. This document makes it simple for potential investors, financial institutions and company management to understand and absorb. Even if you intend to self-finance, a business plan can help you flesh out your idea and spot potential problems. When writing a well-rounded business plan, include the following sections:

• Executive summary: The executive summary should be the first item in the business plan, but it should be written last. It describes the proposed new business and highlights the goals of the company and the methods to achieve them.
• Company description: The company description covers what problems your product or service solves and why your business or idea is best. For example, maybe your background is in molecular engineering, and you’ve used that background to create a new type of athletic wear—you have the proper credentials to make the best material.
• Market analysis: This section of the business plan analyzes how well a company is positioned against its competitors. The market analysis should include target market, segmentation analysis, market size, growth rate, trends and a competitive environment assessment.
• Organization and structure: Write about the type of business organization you expect, what risk management strategies you propose and who will staff the management team. What are their qualifications? Will your business be a single-member limited liability company (LLC) or a corporation ?
• Mission and goals: This section should contain a brief mission statement and detail what the business wishes to accomplish and the steps to get there. These goals should be SMART (specific, measurable, action-orientated, realistic and time-bound).
• Products or services: This section describes how your business will operate. It includes what products you’ll offer to consumers at the beginning of the business, how they compare to existing competitors, how much your products cost, who will be responsible for creating the products, how you’ll source materials and how much they cost to make.
• Background summary: This portion of the business plan is the most time-consuming to write. Compile and summarize any data, articles and research studies on trends that could positively and negatively affect your business or industry.
• Marketing plan: The marketing plan identifies the characteristics of your product or service, summarizes the SWOT analysis and analyzes competitors. It also discusses how you’ll promote your business, how much money will be spent on marketing and how long the campaign is expected to last.
• Financial plan: The financial plan is perhaps the core of the business plan because, without money, the business will not move forward. Include a proposed budget in your financial plan along with projected financial statements, such as an income statement, a balance sheet and a statement of cash flows. Usually, five years of projected financial statements are acceptable. This section is also where you should include your funding request if you’re looking for outside funding.

Learn more: Download our free simple business plan template .

Come Up With an Exit Strategy

An exit strategy is important for any business that is seeking funding because it outlines how you’ll sell the company or transfer ownership if you decide to retire or move on to other projects. An exit strategy also allows you to get the most value out of your business when it’s time to sell. There are a few different options for exiting a business, and the best option for you depends on your goals and circumstances.

The most common exit strategies are:

• Selling the business to another party
• Passing the business down to family members
• Liquidating the business assets
• Closing the doors and walking away

Develop a Scalable Business Model

As your small business grows, it’s important to have a scalable business model so that you can accommodate additional customers without incurring additional costs. A scalable business model is one that can be replicated easily to serve more customers without a significant increase in expenses.

Some common scalable business models are:

• Subscription-based businesses
• Businesses that sell digital products
• Franchise businesses
• Network marketing businesses

Start Planning for Taxes

One of the most important things to do when starting a small business is to start planning for taxes. Taxes can be complex, and there are several different types of taxes you may be liable for, including income tax, self-employment tax, sales tax and property tax. Depending on the type of business you’re operating, you may also be required to pay other taxes, such as payroll tax or unemployment tax.

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When structuring your business, it’s essential to consider how each structure impacts the amount of taxes you owe, daily operations and whether your personal assets are at risk.

An LLC limits your personal liability for business debts. LLCs can be owned by one or more people or companies and must include a registered agent . These owners are referred to as members.

• LLCs offer liability protection for the owners
• They’re one of the easiest business entities to set up
• You can have a single-member LLC
• You may be required to file additional paperwork with your state on a regular basis
• LLCs can’t issue stock
• You’ll need to pay annual filing fees to your state

Limited Liability Partnership (LLP)

An LLP is similar to an LLC but is typically used for licensed business professionals such as an attorney or accountant. These arrangements require a partnership agreement.

• Partners have limited liability for the debts and actions of the LLP
• LLPs are easy to form and don’t require much paperwork
• There’s no limit to the number of partners in an LLP
• Partners are required to actively take part in the business
• LLPs can’t issue stock
• All partners are personally liable for any malpractice claims against the business

Sole Proprietorship

If you start a solo business, you might consider a sole proprietorship . The company and the owner, for legal and tax purposes, are considered the same. The business owner assumes liability for the business. So, if the business fails, the owner is personally and financially responsible for all business debts.

• Sole proprietorships are easy to form
• There’s no need to file additional paperwork with your state
• You’re in complete control of the business
• You’re personally liable for all business debts
• It can be difficult to raise money for a sole proprietorship
• The business may have a limited lifespan

Corporation

A corporation limits your personal liability for business debts just as an LLC does. A corporation can be taxed as a C corporation (C-corp) or an S corporation (S-corp). S-corp status offers pass-through taxation to small corporations that meet certain IRS requirements. Larger companies and startups hoping to attract venture capital are usually taxed as C-corps.

• Corporations offer liability protection for the owners
• The life span of a corporation is not limited
• A corporation can have an unlimited number of shareholders
• Corporations are subject to double taxation
• They’re more expensive and complicated to set up than other business structures
• The shareholders may have limited liability

Before you decide on a business structure, discuss your situation with a small business accountant and possibly an attorney, as each business type has different tax treatments that could affect your bottom line.

Helpful Resources

• How To Set Up an LLC in 7 Steps
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There are several legal issues to address when starting a business after choosing the business structure. The following is a good checklist of items to consider when establishing your business:

Choose Your Business Name

Make it memorable but not too difficult. Choose the same domain name, if available, to establish your internet presence. A business name cannot be the same as another registered company in your state, nor can it infringe on another trademark or service mark that is already registered with the United States Patent and Trademark Office (USPTO).

Business Name vs. DBA

There are business names, and then there are fictitious business names known as “Doing Business As” or DBA. You may need to file a DBA if you’re operating under a name that’s different from the legal name of your business. For example, “Mike’s Bike Shop” is doing business as “Mike’s Bikes.” The legal name of the business is “Mike’s Bike Shop,” and “Mike’s Bikes” is the DBA.

You may need to file a DBA with your state, county or city government offices. The benefits of a DBA include:

• It can help you open a business bank account under your business name
• A DBA can be used as a “trade name” to brand your products or services
• A DBA can be used to get a business license

Register Your Business and Obtain an EIN

You’ll officially create a corporation, LLC or other business entity by filing forms with your state’s business agency―usually the Secretary of State. As part of this process, you’ll need to choose a registered agent to accept legal documents on behalf of your business. You’ll also pay a filing fee. The state will send you a certificate that you can use to apply for licenses, a tax identification number (TIN) and business bank accounts.

Next, apply for an employer identification number (EIN) . All businesses, other than sole proprietorships with no employees, must have a federal employer identification number. Submit your application to the IRS and you’ll typically receive your number in minutes.

Get Appropriate Licenses and Permits

Legal requirements are determined by your industry and jurisdiction. Most businesses need a mixture of local, state and federal licenses to operate. Check with your local government office (and even an attorney) for licensing information tailored to your area.

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Open a Business Bank Account

Keep your business and personal finances separate. Here’s how to choose a business checking account —and why separate business accounts are essential. When you open a business bank account, you’ll need to provide your business name and your business tax identification number (EIN). This business bank account can be used for your business transactions, such as paying suppliers or invoicing customers. Most times, a bank will require a separate business bank account to issue a business loan or line of credit.

Hire a Bookkeeper or Get Accounting Software

If you sell a product, you need an inventory function in your accounting software to manage and track inventory. The software should have ledger and journal entries and the ability to generate financial statements.

Some software programs double as bookkeeping tools. These often include features such as check writing and managing receivables and payables. You can also use this software to track your income and expenses, generate invoices, run reports and calculate taxes.

There are many bookkeeping services available that can do all of this for you, and more. These services can be accessed online from any computer or mobile device and often include features such as bank reconciliation and invoicing. Check out the best accounting software for small business, or see if you want to handle the bookkeeping yourself.

Determine Your Break-Even Point

Before you fund your business, you must get an idea of your startup costs. To determine these, make a list of all the physical supplies you need, estimate the cost of any professional services you will require, determine the price of any licenses or permits required to operate and calculate the cost of office space or other real estate. Add in the costs of payroll and benefits, if applicable.

Businesses can take years to turn a profit, so it’s better to overestimate the startup costs and have too much money than too little. Many experts recommend having enough cash on hand to cover six months of operating expenses.

When you know how much you need to get started with your business, you need to know the point at which your business makes money. This figure is your break-even point.

In contrast, the contribution margin = total sales revenue – cost to make product

For example, let’s say you’re starting a small business that sells miniature birdhouses for fairy gardens. You have determined that it will cost you $500 in startup costs. Your variable costs are $0.40 per birdhouse produced, and you sell them for $1.50 each.

Let’s write these out so it’s easy to follow:

This means that you need to sell at least 456 units just to cover your costs. If you can sell more than 456 units in your first month, you will make a profit.

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There are many different ways to fund your business—some require considerable effort, while others are easier to obtain. Two categories of funding exist: internal and external.

Internal funding includes:

• Personal savings
• Credit cards
• Funds from friends and family

If you finance the business with your own funds or with credit cards, you have to pay the debt on the credit cards and you’ve lost a chunk of your wealth if the business fails. By allowing your family members or friends to invest in your business, you are risking hard feelings and strained relationships if the company goes under. Business owners who want to minimize these risks may consider external funding.

External funding includes:

• Small business loans
• Small business grants
• Angel investors
• Venture capital
• Crowdfunding

Small businesses may have to use a combination of several sources of capital. Consider how much money is needed, how long it will take before the company can repay it and how risk-tolerant you are. No matter which source you use, plan for profit. It’s far better to take home six figures than make seven figures and only keep $80,000 of it.

Funding ideas include:

• Invoice factoring: With invoice factoring , you can sell your unpaid invoices to a third party at a discount.
• Business lines of credit: Apply for a business line of credit , which is similar to a personal line of credit. The credit limit and interest rate will be based on your business’s revenue, credit score and financial history.
• Equipment financing: If you need to purchase expensive equipment for your business, you can finance it with a loan or lease.
• Small Business Administration (SBA) microloans: Microloans are up to $50,000 loans that can be used for working capital, inventory or supplies and machinery or equipment.
• Grants: The federal government offers grants for businesses that promote innovation, export growth or are located in historically disadvantaged areas. You can also find grants through local and regional organizations.
• Crowdfunding: With crowdfunding , you can raise money from a large group of people by soliciting donations or selling equity in your company.

Choose the right funding source for your business by considering the amount of money you need, the time frame for repayment and your tolerance for risk.

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You need to have insurance for your business , even if it’s a home-based business or you don’t have any employees. The type of insurance you need depends on your business model and what risks you face. You might need more than one type of policy, and you might need additional coverage as your business grows. In most states, workers’ compensation insurance is required by law if you have employees.

Work With an Agent To Get Insured

An insurance agent can help determine what coverages are appropriate for your business and find policies from insurers that offer the best rates. An independent insurance agent represents several different insurers, so they can shop around for the best rates and coverage options.

Basic Types of Business Insurance Coverage

• Liability insurance protects your business against third-party claims of bodily injury, property damage and personal injury such as defamation or false advertising.
• Property insurance covers the physical assets of your business, including your office space, equipment and inventory.
• Business interruption insurance pays for the loss of income if your business is forced to close temporarily due to a covered event such as a natural disaster.
• Product liability insurance protects against claims that your products caused bodily injury or property damage.
• Employee practices liability insurance covers claims from employees alleging discrimination, sexual harassment or other wrongful termination.
• Workers’ compensation insurance covers medical expenses and income replacement for employees who are injured on the job.
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Business tools can help make your life easier and make your business run more smoothly. The right tools can help you save time, automate tasks and make better decisions.

Consider the following tools in your arsenal:

• Accounting software : Track your business income and expenses, prepare financial statements and file taxes. Examples include QuickBooks and FreshBooks.
• Customer relationship management (CRM) software : This will help you manage your customer relationships, track sales and marketing data and automate tasks like customer service and follow-ups. Examples include Zoho CRM and monday.com.
• Project management software : Plan, execute and track projects. It can also be used to manage employee tasks and allocate resources. Examples include Airtable and ClickUp.
• Credit card processor : This will allow you to accept credit card payments from customers. Examples include Stripe and PayPal.
• Point of sale (POS) : A system that allows you to process customer payments. Some accounting software and CRM software have POS features built-in. Examples include Clover and Lightspeed.
• Virtual private network (VPN) : Provides a secure, private connection between your computer and the internet. This is important for businesses that handle sensitive data. Examples include NordVPN and ExpressVPN.
• Merchant services : When customers make a purchase, the money is deposited into your business account. You can also use merchant services to set up recurring billing or subscription payments. Examples include Square and Stripe.
• Email hosting : This allows you to create a professional email address with your own domain name. Examples include G Suite and Microsoft Office 365.

Many business owners spend so much money creating their products that there isn’t a marketing budget by the time they’ve launched. Alternatively, they’ve spent so much time developing the product that marketing is an afterthought.

Create a Website

Even if you’re a brick-and-mortar business, a web presence is essential. Creating a website doesn’t take long, either—you can have one done in as little as a weekend. You can make a standard informational website or an e-commerce site where you sell products online. If you sell products or services offline, include a page on your site where customers can find your locations and hours. Other pages to add include an “About Us” page, product or service pages, frequently asked questions (FAQs), a blog and contact information.

Optimize Your Site for SEO

After getting a website or e-commerce store, focus on optimizing it for search engines (SEO). This way, when a potential customer searches for specific keywords for your products, the search engine can point them to your site. SEO is a long-term strategy, so don’t expect a ton of traffic from search engines initially—even if you’re using all the right keywords.

Create Relevant Content

Provide quality digital content on your site that makes it easy for customers to find the correct answers to their questions. Content marketing ideas include videos, customer testimonials, blog posts and demos. Consider content marketing one of the most critical tasks on your daily to-do list. This is used in conjunction with posting on social media.

Get Listed in Online Directories

Customers use online directories like Yelp, Google My Business and Facebook to find local businesses. Some city halls and chambers of commerce have business directories too. Include your business in as many relevant directories as possible. You can also create listings for your business on specific directories that focus on your industry.

Develop a Social Media Strategy

Your potential customers are using social media every day—you need to be there too. Post content that’s interesting and relevant to your audience. Use social media to drive traffic back to your website where customers can learn more about what you do and buy your products or services.

You don’t necessarily need to be on every social media platform available. However, you should have a presence on Facebook and Instagram because they offer e-commerce features that allow you to sell directly from your social media accounts. Both of these platforms have free ad training to help you market your business.

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To scale your business, you need to grow your customer base and revenue. This can be done by expanding your marketing efforts, improving your product or service, collaborating with other creators or adding new products or services that complement what you already offer.

Think about ways you can automate or outsource certain tasks so you can focus on scaling the business. For example, if social media marketing is taking up too much of your time, consider using a platform such as Hootsuite to help you manage your accounts more efficiently. You can also consider outsourcing the time-consumer completely.

You can also use technology to automate certain business processes, including accounting, email marketing and lead generation. Doing this will give you more time to focus on other aspects of your business.

When scaling your business, it’s important to keep an eye on your finances and make sure you’re still profitable. If you’re not making enough money to cover your costs, you need to either reduce your expenses or find ways to increase your revenue.

Build a Team

As your business grows, you’ll need to delegate tasks and put together a team of people who can help you run the day-to-day operations. This might include hiring additional staff, contractors or freelancers.

Resources for building a team include:

• Hiring platforms: To find the right candidates, hiring platforms, such as Indeed and Glassdoor, can help you post job descriptions, screen résumés and conduct video interviews.
• Job boards: Job boards such as Craigslist and Indeed allow you to post open positions for free.
• Social media: You can also use social media platforms such as LinkedIn and Facebook to find potential employees.
• Freelance platforms: Using Upwork, Freelancer and Fiverr can help you find talented freelancers for one-time or short-term projects. You can also outsource certain tasks, such as customer service, social media marketing or bookkeeping.

You might also consider partnering with other businesses in your industry. For example, if you’re a wedding planner, you could partner with a florist, photographer, catering company or venue. This way, you can offer your customers a one-stop shop for all their wedding needs. Another example is an e-commerce store that partners with a fulfillment center. This type of partnership can help you save money on shipping and storage costs, and it can also help you get your products to your customers faster.

To find potential partnerships, search for businesses in your industry that complement what you do. For example, if you’re a web designer, you could partner with a digital marketing agency.

You can also search for businesses that serve the same target market as you but offer different products or services. For example, if you sell women’s clothing, you could partner with a jewelry store or a hair salon.

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To rank the best states to start a business in 2024, Forbes Advisor analyzed 18 key metrics across five categories to determine which states are the best and worst to start a business in. Our ranking takes into consideration factors that impact businesses and their ability to succeed, such as business costs, business climate, economy, workforce and financial accessibility in each state. Check out the full report .

Starting a small business takes time, effort and perseverance. But if you’re willing to put in the work, it can be a great way to achieve your dreams and goals. Be sure to do your research, create a solid business plan and pivot along the way. Once you’re operational, don’t forget to stay focused and organized so you can continue to grow your business.

How do I start a small business with no money?

There are several funding sources for brand-new businesses and most require a business plan to secure it. These include the SBA , private grants, angel investors, crowdfunding and venture capital.

What is the best business structure?

The best business structure for your business will depend entirely on what kind of company you form, your industry and what you want to accomplish. But any successful business structure will be one that will help your company set realistic goals and follow through on set tasks.

Do I need a business credit card?

You don’t need one, but a business credit card can be helpful for new small businesses. It allows you to start building business credit, which can help you down the road when you need to take out a loan or line of credit. Additionally, business credit cards often come with rewards and perks that can save you money on business expenses.

Do I need a special license or permit to start a small business?

The answer to this question will depend on the type of business you want to start and where you’re located. Some businesses, such as restaurants, will require a special permit or license to operate. Others, such as home daycare providers, may need to register with the state.

How much does it cost to create a business?

The cost of starting a business will vary depending on the size and type of company you want to create. For example, a home-based business will be less expensive to start than a brick-and-mortar store. Additionally, the cost of starting a business will increase if you need to rent or buy commercial space, hire employees or purchase inventory. You could potentially get started for free by dropshipping or selling digital goods.

How do I get a loan for a new business?

The best way to get a loan for a new business is to approach banks or other financial institutions and provide them with a business plan and your financial history. You can also look into government-backed loans, such as those offered by the SBA. Startups may also be able to get loans from alternative lenders, including online platforms such as Kiva.

Do I need a business degree to start a business?

No, you don’t need a business degree to start a business. However, acquiring a degree in business or a related field can provide you with the understanding and ability to run an effective company. Additionally, you may want to consider taking some business courses if you don’t have a degree to learn more about starting and running a business. You can find these online and at your local Small Business Administration office.

What are some easy businesses to start?

One of the easiest businesses to start also has the lowest overhead: selling digital goods. This can include items such as e-books, online courses, audio files or software. If you have expertise in a particular area or niche, this is a great option for you. Dropshipping is also a great option because you don’t have to keep inventory. You could also buy wholesale products or create your own. Once you create your product, you can sell it through your own website or third-party platforms such as Amazon or Etsy.

What is the most profitable type of business?

There is no one answer to this question because the most profitable type of business will vary depending on a number of factors, such as your industry, location, target market and business model. However, some businesses tend to be more profitable than others, such as luxury goods, high-end services, business-to-business companies and subscription-based businesses. If you’re not sure what type of business to start, consider your strengths and interests, as well as the needs of your target market, to help you choose a profitable business idea.

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Katherine Haan is a small business owner with nearly two decades of experience helping other business owners increase their incomes.

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