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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 (B.S., M.Ed.) of Calcworkshop® teaching initial value problems

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 7: Second-Order Differential Equations

Initial-value problems and boundary-value problems, learning objectives.

Solve initial-value and boundary-value problems involving linear differential equations.

So far, we have been finding general solutions to differential equations. However, differential equations are often used to describe physical systems, and the person studying that physical system usually knows something about the state of that system at one or more points in time. For example, if a constant-coefficient differential equation is representing how far a motorcycle shock absorber is compressed, we might know that the rider is sitting still on his motorcycle at the start of a race, time [latex]t=t_0[/latex]. This means the system is at equilibrium, so [latex]y(t_0)=0[/latex], and the compression of the shock absorber is not changing, so [latex]y'(t_0)=0[/latex]. With these two initial conditions and the general solution to the differential equation, we can find the specific solution to the differential equation that satisfies both initial conditions. This process is known as solving an initial-value problem . (Recall that we discussed initial-value problems in Introduction to Differential Equations .) Note that second-order equations have two arbitrary constants in the general solution, and therefore we require two initial conditions to find the solution to the initial-value problem.

Sometimes we know the condition of the system at two different times. For example, we might know [latex]y(t_0)=y_0[/latex] and [latex]y(t_1)=y_1[/latex]. These conditions are called boundary conditions , and finding the solution to the differential equation that satisfies the boundary conditions is called solving a boundary-value problem .

Mathematicians, scientists, and engineers are interested in understanding the conditions under which an initial-value problem or a boundary-value problem has a unique solution. Although a complete treatment of this topic is beyond the scope of this text, it is useful to know that, within the context of constant-coefficient, second-order equations, initial-value problems are guaranteed to have a unique solution as long as two initial conditions are provided. Boundary-value problems, however, are not as well behaved. Even when two boundary conditions are known, we may encounter boundary-value problems with unique solutions, many solutions, or no solution at all.

Example: solving an initial-value problem

Solve the following initial-value problem: [latex]y''+3y'-4y=0[/latex], [latex]y(0)=1[/latex], [latex]y'(0)=-9[/latex].

We already solved this differential equation in Example “Solving Second-Order Equations with Constant Coefficients” part a. and found the general solution to be

[latex]y(x)=c_1e^{-4x}+c_2e^x[/latex].

[latex]y^\prime=-4c_1e^{-4x}+c_2e^x[/latex].

When [latex]x=0[/latex], we have [latex]y(0)=c_1+c_2[/latex] and [latex]y^\prime(0)=-4c_1+c_2[/latex]. Applying the initial conditions, we have

[latex]\begin{aligned} c_1+c_2&=1 \\ -4c_1+c_2&=-9 \end{aligned}[/latex].

Then [latex]c_1=1-c_2[/latex]. Substituting this expression into the second equation, we see that

[latex]\begin{aligned} -4(1-c_2)+c_2&=-9 \\ -4+4c_2+c_2&=-9 \\ 5c_2&=-5 \\ c_2&=-1 \end{aligned}[/latex].

So, [latex]c_1=2[/latex] and the solution to the initial-value problem is

[latex]y(x)=2e^{-4x}-e^x[/latex].

Solve the initial-value problem [latex]y''-3y'-10y=0[/latex], [latex]y(0)=0[/latex], [latex]y'(0)=7[/latex].

[latex]y(x)=-e^{-2x}+e^{-5x}[/latex].

Watch the following video to see the worked solution to the above Try It

Example: solving an initial-value problem and graphing the solution

Solve the following initial-value problem and graph the solution:

[latex]y^{\prime\prime}+6y^\prime+13y=0, \ y(0)=0, \ y^\prime(0)=2[/latex].

We already solved this differential equation in Example “Solving Second-Order Equations with Constant Coefficients” part b. and found the general solution to be

[latex]y(x)=e^{-3x}(c_1\cos2x+c_2\sin2x)[/latex].

[latex]y^\prime(x)=e^{-3x}(-2c_1\sin2x+2c_2\cos2x)-3e^{-3x}(c_1\cos2x+c_2\sin2x)[/latex].

When [latex]x=0[/latex], we have [latex]y(0)=c_1[/latex] and [latex]y^\prime(0)=2c_2-3c_1[/latex]. Applying the initial conditions, we obtain

[latex]\begin{aligned} c_1&=0 \\ -3c_1+2c_2&= 2 \end{aligned}[/latex].

Therefore, [latex]c_1=0[/latex], [latex]c_2=1[/latex], and the solution to the initial value problem is shown in the following graph.

[latex]y=e^{-3x}\sin2x[/latex]

This figure is a graph of the function y = e^−3x sin 2x. The x axis is scaled in increments of tenths. The y axis is scaled in increments of even tenths. The curve passes through the origin and has a horizontal asymptote of the positive x axis.

Solve the following initial-value problem and graph the solution: [latex]y''-2y'+10y=10=0[/latex], [latex]y(0)=2[/latex], [latex]y'(0)=-1[/latex].

[latex]y(x)=e^{x}(2\cos3x-\sin3x)[/latex]

This figure is the graph of y(x) = e^x(2 cos 3x − sin 3x) It has the positive x axis scaled in increments of even tenths. The y axis is scaled in increments of twenty. The graph itself starts at the origin. Its amplitude increases as x increases.

Figure 2. Graph of [latex]y(x)=e^{x}(2\cos3x-\sin3x)[/latex].

Example: initial-value problem representing a spring-mass system

The following initial-value problem models the position of an object with mass attached to a spring. Spring-mass systems are examined in detail in Applications . The solution to the differential equation gives the position of the mass with respect to a neutral (equilibrium) position (in meters) at any given time. (Note that for spring-mass systems of this type, it is customary to define the downward direction as positive.)

[latex]y^{\prime\prime}+2y^\prime+y=0, \ y(0)=1, \ y^\prime(0)=0[/latex]

Solve the initial-value problem and graph the solution. What is the position of the mass at time [latex]t=2[/latex] sec? How fast is the mass moving at time [latex]t=1[/latex] sec? In what direction?

In Example “Solving Second-Order Equations with Constant Coefficients” part c. we found the general solution to this differential equation to be

[latex]y(t)=c_1e^{-t}+c_2te^{-t}[/latex].

[latex]y^\prime(t)=-c_1e^{-t}+c_2(-te^{-t}+e^{-t})[/latex].

When [latex]t=0[/latex], we have [latex]y(0)=c_1[/latex] and [latex]y'(0)=-c_1+c_2[/latex]. Applying the initial conditions, we obtain

[latex]\begin{aligned} c_1&=1 \\ -c_1+c_2&= 0 \end{aligned}[/latex].

Thus, [latex]c_1=1[/latex], [latex]c_2=1[/latex], and the solution to the initial value problem is

[latex]y(t)=e^{-t}+te^{-t}[/latex].

This solution is represented in the following graph. At time [latex]t=2[/latex], the mass is at position [latex]y(2)=e^{-2}+2e^{-2}=3e^{-2}\approx0.406[/latex] [latex]m[/latex] below equilibrium.

This figure is the graph of y(t) = e^−t + te^−t. The horizontal axis is labeled with t and is scaled in increments of even tenths. The y axis is scaled in increments of 0.5. The graph passes through positive one and decreases with a horizontal asymptote of the positive t axis.

To calculate the velocity at time [latex]t=1[/latex], we need to find the derivative. We have [latex]y(t)=e^{-t}+te^{-t}[/latex], so

[latex]y^\prime(t)=-e^{-t}+e^{-t}-te^{-t}=-te^{-t}[/latex].

Then [latex]y^\prime(1)=e^{-1}\approx-0.3679[/latex]. At time [latex]t=1[/latex], the mass is moving upward at [latex]0.3679[/latex] m/sec.

Suppose the following initial-value problem models the position (in feet) of a mass in a spring-mass system at any given time. Solve the initial-value problem and graph the solution. What is the position of the mass at time [latex]t=0.3[/latex] sec? How fast is it moving at time [latex]t=0.1[/latex] sec? In what direction?

[latex]y^{\prime\prime}+14y^\prime+49y=0, \ y(0)=0, \ y^\prime(0)=1[/latex]

[latex]y(t)=te^{-7t}[/latex]

This figure is the graph of y(t) = te^−7t. The horizontal axis is labeled with t and is scaled in increments of tenths. The y axis is scaled in increments of 0.5. The graph passes through the origin and has a horizontal asymptote of the positive t axis.

At time [latex]t=0.3[/latex], [latex]y(0.3)=0.3^{(-7*0.3)}=0.3e^{-2.1}\approx0.0367[/latex]. The mass is [latex]0.0367[/latex] ft below equilibrium. At time [latex]t=0.1[/latex], [latex]y^\prime(0.1)=0.3e^{-0.7}\approx0.1490[/latex]. The mass is moving downward at a speed of [latex]0.1490[/latex] ft/sec.

Example: solving a boundary-value problem

In Example “Solving Second-Order Equations with Constant Coefficients” part f. we solved the differential equation [latex]y''+16y=0[/latex] and found the general solution to be [latex]y(t)=c_1\cos4t+c_2\sin4t[/latex]. If possible, solve the boundary-value problem if the boundary conditions are the following:

[latex]y(0)=0[/latex], [latex]y\left(\frac{\pi}4\right)=0[/latex]
[latex]y(0)=1[/latex], [latex]y\left(\frac{\pi}8\right)=0[/latex]
[latex]y\left(\frac{\pi}8\right)=0[/latex], [latex]y\left(\frac{3\pi}8\right)=0[/latex]

[latex]y(x)=c_1\cos{4t}+c_2\sin{4t}[/latex]

Applying the first boundary condition given here, we get [latex]y(0)=c_1=0[/latex]. So the solution is of the form [latex]y(t)=c_2\sin4t[/latex]. When we apply the second boundary condition, though, we get [latex]y\left(\frac{\pi}4\right)=c_2\sin\left(4\left(\frac{\pi}4\right)\right)=c_2\sin\pi=0[/latex] for all values of [latex]c_2[/latex]. The boundary conditions are not sufficient to determine a value for [latex]c_2[/latex], so this boundary-value problem has infinitely many solutions. Thus, [latex]y(t)=c_2\sin4t[/latex] is a solution for any value of [latex]c_2[/latex].
Applying the first boundary condition given here, we get [latex]y(0)=c_1=1[/latex]. Applying the second boundary condition gives[latex]y(\frac{\pi}{8})=c_2=0[/latex], so [latex]c_2=0[/latex]. In this case, we have a unique solution: [latex]y(t)=\cos 4t[/latex].
Applying the first boundary condition given here, we get [latex]y\left(\frac{\pi}8\right)=c_2=0[/latex]. However, applying the second boundary condition gives [latex]y\left(\frac{3\pi}8\right)=-c_2=2[/latex], so [latex]c_2=-2[/latex]. We cannot have [latex]c_2=0=-2[/latex] so this boundary value problem has no solution.
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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.

Solve Initial Value Problem Definition Application and

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) .

Generic example of solving initial value problem

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) .

initial value problem solution y equals 2 times exponential power minus 2 times t

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) .

initial value problem solution y equals sin2t

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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]

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8.1: Basics of Differential Equations

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Learning Objectives

Identify the order of a differential equation.
Explain what is meant by a solution to a differential equation.
Distinguish between the general solution and a particular solution of a differential equation.
Identify an initial-value problem.
Identify whether a given function is a solution to a differential equation or an initial-value problem.

Calculus is the mathematics of change, and rates of change are expressed by derivatives. Thus, one of the most common ways to use calculus is to set up an equation containing an unknown function $y=f(x)$ and its derivative, known as a differential equation. Solving such equations often provides information about how quantities change and frequently provides insight into how and why the changes occur.

Techniques for solving differential equations can take many different forms, including direct solution, use of graphs, or computer calculations. We introduce the main ideas in this chapter and describe them in a little more detail later in the course. In this section we study what differential equations are, how to verify their solutions, some methods that are used for solving them, and some examples of common and useful equations.

General Differential Equations

Consider the equation $y′=3x^2,$ which is an example of a differential equation because it includes a derivative. There is a relationship between the variables $x$ and $y:y$ is an unknown function of $x$. Furthermore, the left-hand side of the equation is the derivative of $y$. Therefore we can interpret this equation as follows: Start with some function $y=f(x)$ and take its derivative. The answer must be equal to $3x^2$. What function has a derivative that is equal to $3x^2$? One such function is $y=x^3$, so this function is considered a solution to a differential equation.

Definition: differential equation

A differential equation is an equation involving an unknown function $y=f(x)$ and one or more of its derivatives. A solution to a differential equation is a function $y=f(x)$ that satisfies the differential equation when $f$ and its derivatives are substituted into the equation.

Go to this website to explore more on this topic.

Some examples of differential equations and their solutions appear in Table $\PageIndex{1}$.

Note that a solution to a differential equation is not necessarily unique, primarily because the derivative of a constant is zero. For example, $y=x^2+4$ is also a solution to the first differential equation in Table $\PageIndex{1}$. We will return to this idea a little bit later in this section. For now, let’s focus on what it means for a function to be a solution to a differential equation.

Example $\PageIndex{1}$: Verifying Solutions of Differential Equations

Verify that the function $y=e^{−3x}+2x+3$ is a solution to the differential equation $y′+3y=6x+11$.

To verify the solution, we first calculate $y′$ using the chain rule for derivatives. This gives $y′=−3e^{−3x}+2$. Next we substitute $y$ and $y′$ into the left-hand side of the differential equation:

$(−3e^{−2x}+2)+3(e^{−2x}+2x+3).$

The resulting expression can be simplified by first distributing to eliminate the parentheses, giving

$−3e^{−2x}+2+3e^{−2x}+6x+9.$

Combining like terms leads to the expression $6x+11$, which is equal to the right-hand side of the differential equation. This result verifies that $y=e^{−3x}+2x+3$ is a solution of the differential equation.

Exercise $\PageIndex{1}$

Verify that $y=2e^{3x}−2x−2$ is a solution to the differential equation $y′−3y=6x+4.$

First calculate $y′$ then substitute both $y′$ and $y$ into the left-hand side.

It is convenient to define characteristics of differential equations that make it easier to talk about them and categorize them. The most basic characteristic of a differential equation is its order.

Definition: order of a differential equation

The order of a differential equation is the highest order of any derivative of the unknown function that appears in the equation.

Example $\PageIndex{2}$: Identifying the Order of a Differential Equation

The highest derivative in the equation is $y′$,

What is the order of each of the following differential equations?

$y′−4y=x^2−3x+4$
$x^2y'''−3xy''+xy′−3y=\sin x$
$\frac{4}{x}y^{(4)}−\frac{6}{x^2}y''+\frac{12}{x^4}y=x^3−3x^2+4x−12$
The highest derivative in the equation is $y′$,so the order is $1$.
The highest derivative in the equation is $y'''$, so the order is $3$.
The highest derivative in the equation is $y^{(4)}$, so the order is $4$.

Exercise $\PageIndex{2}$

What is the order of the following differential equation?

$(x^4−3x)y^{(5)}−(3x^2+1)y′+3y=\sin x\cos x$

What is the highest derivative in the equation?

General and Particular Solutions

We already noted that the differential equation $y′=2x$ has at least two solutions: $y=x^2$ and $y=x^2+4$. The only difference between these two solutions is the last term, which is a constant. What if the last term is a different constant? Will this expression still be a solution to the differential equation? In fact, any function of the form $y=x^2+C$, where $C$ represents any constant, is a solution as well. The reason is that the derivative of $x^2+C$ is $2x$, regardless of the value of $C$. It can be shown that any solution of this differential equation must be of the form $y=x^2+C$. This is an example of a general solution to a differential equation. A graph of some of these solutions is given in Figure $\PageIndex{1}$. (Note: in this graph we used even integer values for C ranging between $−4$ and $4$. In fact, there is no restriction on the value of $C$; it can be an integer or not.)

$A graph of a family of solutions to the differential equation y’ = 2 x, which are of the form y = x ^ 2 + C. Parabolas are drawn for values of C: -4, -2, 0, 2, and 4.$

In this example, we are free to choose any solution we wish; for example, $y=x^2−3$ is a member of the family of solutions to this differential equation. This is called a particular solution to the differential equation. A particular solution can often be uniquely identified if we are given additional information about the problem.

Example $\PageIndex{3}$: Finding a Particular Solution

Find the particular solution to the differential equation $y′=2x$ passing through the point $(2,7)$.

Any function of the form $y=x^2+C$ is a solution to this differential equation. To determine the value of $C$, we substitute the values $x=2$ and $y=7$ into this equation and solve for $C$:

\[ \begin{align*} y =x^2+C \\[4pt] 7 =2^2+C \\[4pt] =4+C \\[4pt] C =3. \end{align*}\]

Therefore the particular solution passing through the point $(2,7)$ is $y=x^2+3$.

Exercise $\PageIndex{3}$

Find the particular solution to the differential equation

\[ y′=4x+3 \nonumber \]

passing through the point $(1,7),$ given that $y=2x^2+3x+C$ is a general solution to the differential equation.

First substitute $x=1$ and $y=7$ into the equation, then solve for $C$.

\[ y=2x^2+3x+2 \nonumber \]

Initial-Value Problems

Usually a given differential equation has an infinite number of solutions, so it is natural to ask which one we want to use. To choose one solution, more information is needed. Some specific information that can be useful is an initial value , which is an ordered pair that is used to find a particular solution.

A differential equation together with one or more initial values is called an initial-value problem . The general rule is that the number of initial values needed for an initial-value problem is equal to the order of the differential equation. For example, if we have the differential equation $y′=2x$, then $y(3)=7$ is an initial value, and when taken together, these equations form an initial-value problem. The differential equation $y''−3y′+2y=4e^x$ is second order, so we need two initial values. With initial-value problems of order greater than one, the same value should be used for the independent variable. An example of initial values for this second-order equation would be $y(0)=2$ and $y′(0)=−1.$ These two initial values together with the differential equation form an initial-value problem. These problems are so named because often the independent variable in the unknown function is $t$, which represents time. Thus, a value of $t=0$ represents the beginning of the problem.

Example $\PageIndex{4}$: Verifying a Solution to an Initial-Value Problem

Verify that the function $y=2e^{−2t}+e^t$ is a solution to the initial-value problem

\[ y′+2y=3e^t, \quad y(0)=3.\nonumber \]

For a function to satisfy an initial-value problem, it must satisfy both the differential equation and the initial condition. To show that $y$ satisfies the differential equation, we start by calculating $y′$. This gives $y′=−4e^{−2t}+e^t$. Next we substitute both $y$ and $y′$ into the left-hand side of the differential equation and simplify:

\[ \begin{align*} y′+2y &=(−4e^{−2t}+e^t)+2(2e^{−2t}+e^t) \\[4pt] &=−4e^{−2t}+e^t+4e^{−2t}+2e^t =3e^t. \end{align*}\]

This is equal to the right-hand side of the differential equation, so $y=2e^{−2t}+e^t$ solves the differential equation. Next we calculate $y(0)$:

\[ y(0)=2e^{−2(0)}+e^0=2+1=3. \nonumber \]

This result verifies the initial value. Therefore the given function satisfies the initial-value problem.

Exercise $\PageIndex{4}$

Verify that $y=3e^{2t}+4\sin t$ is a solution to the initial-value problem

\[ y′−2y=4\cos t−8\sin t,y(0)=3. \nonumber \]

First verify that $y$ solves the differential equation. Then check the initial value.

In Example $\PageIndex{4}$, the initial-value problem consisted of two parts. The first part was the differential equation $y′+2y=3e^x$, and the second part was the initial value $y(0)=3.$ These two equations together formed the initial-value problem.

The same is true in general. An initial-value problem will consists of two parts: the differential equation and the initial condition. The differential equation has a family of solutions, and the initial condition determines the value of $C$. The family of solutions to the differential equation in Example $\PageIndex{4}$ is given by $y=2e^{−2t}+Ce^t.$ This family of solutions is shown in Figure $\PageIndex{2}$, with the particular solution $y=2e^{−2t}+e^t$ labeled.

$A graph of a family of solutions to the differential equation y’ + 2 y = 3 e ^ t, which are of the form y = 2 e ^ (-2 t) + C e ^ t. The versions with C = 1, 0.5, and -0.2 are shown, among others not labeled. For all values of C, the function increases rapidly for t < 0 as t goes to negative infinity. For C > 0, the function changes direction and increases in a gentle curve as t goes to infinity. Larger values of C have a tighter curve closer to the y axis and at a higher y value. For C = 0, the function goes to 0 as t goes to infinity. For C < 0, the function continues to decrease as t goes to infinity.$

Example $\PageIndex{5}$: Solving an Initial-value Problem

Solve the following initial-value problem:

\[ y′=3e^x+x^2−4,y(0)=5. \nonumber \]

The first step in solving this initial-value problem is to find a general family of solutions. To do this, we find an antiderivative of both sides of the differential equation

\[∫y′\,dx=∫(3e^x+x^2−4)\,dx, \nonumber \]

$y+C_1=3e^x+\frac{1}{3}x^3−4x+C_2$.

We are able to integrate both sides because the y term appears by itself. Notice that there are two integration constants: $C_1$ and $C_2$. Solving this equation for $y$ gives

$y=3e^x+\frac{1}{3}x^3−4x+C_2−C_1.$

Because $C_1$ and $C_2$ are both constants, $C_2−C_1$ is also a constant. We can therefore define $C=C_2−C_1,$ which leads to the equation

$y=3e^x+\frac{1}{3}x^3−4x+C.$

Next we determine the value of $C$. To do this, we substitute $x=0$ and $y=5$ into this equation and solve for $C$:

\[ \begin{align*} 5 &=3e^0+\frac{1}{3}0^3−4(0)+C \\[4pt] 5 &=3+C \\[4pt] C&=2 \end{align*}. \nonumber \]

Now we substitute the value $C=2$ into the general equation. The solution to the initial-value problem is $y=3e^x+\frac{1}{3}x^3−4x+2.$

The difference between a general solution and a particular solution is that a general solution involves a family of functions, either explicitly or implicitly defined, of the independent variable. The initial value or values determine which particular solution in the family of solutions satisfies the desired conditions.

Exercise $\PageIndex{5}$

Solve the initial-value problem

\[ y′=x^2−4x+3−6e^x,y(0)=8. \nonumber \]

First take the antiderivative of both sides of the differential equation. Then substitute $x=0$ and $y=8$ into the resulting equation and solve for $C$.

$y=\frac{1}{3}x^3−2x^2+3x−6e^x+14$

In physics and engineering applications, we often consider the forces acting upon an object, and use this information to understand the resulting motion that may occur. For example, if we start with an object at Earth’s surface, the primary force acting upon that object is gravity. Physicists and engineers can use this information, along with Newton’s second law of motion (in equation form $F=ma$, where $F$ represents force, $m$ represents mass, and $a$ represents acceleration), to derive an equation that can be solved.

$A picture of a baseball with an arrow underneath it pointing down. The arrow is labeled g = -9.8 m/sec ^ 2.$

In Figure $\PageIndex{3}$ we assume that the only force acting on a baseball is the force of gravity. This assumption ignores air resistance. (The force due to air resistance is considered in a later discussion.) The acceleration due to gravity at Earth’s surface, g, is approximately $9.8\,\text{m/s}^2$. We introduce a frame of reference, where Earth’s surface is at a height of 0 meters. Let $v(t)$ represent the velocity of the object in meters per second. If $v(t)>0$, the ball is rising, and if $v(t)<0$, the ball is falling (Figure).

$A picture of a baseball with an arrow above it pointing up and an arrow below it pointing down. The arrow pointing up is labeled v(t) > 0, and the arrow pointing down is labeled v(t) < 0.$

Our goal is to solve for the velocity $v(t)$ at any time $t$. To do this, we set up an initial-value problem. Suppose the mass of the ball is $m$, where $m$ is measured in kilograms. We use Newton’s second law, which states that the force acting on an object is equal to its mass times its acceleration $(F=ma)$. Acceleration is the derivative of velocity, so $a(t)=v′(t)$. Therefore the force acting on the baseball is given by $F=mv′(t)$. However, this force must be equal to the force of gravity acting on the object, which (again using Newton’s second law) is given by $F_g=−mg$, since this force acts in a downward direction. Therefore we obtain the equation $F=F_g$, which becomes $mv′(t)=−mg$. Dividing both sides of the equation by $m$ gives the equation

\[ v′(t)=−g. \nonumber \]

Notice that this differential equation remains the same regardless of the mass of the object.

We now need an initial value. Because we are solving for velocity, it makes sense in the context of the problem to assume that we know the initial velocity , or the velocity at time $t=0.$ This is denoted by $v(0)=v_0.$

Example $\PageIndex{6}$: Velocity of a Moving Baseball

A baseball is thrown upward from a height of $3$ meters above Earth’s surface with an initial velocity of $10$ m/s, and the only force acting on it is gravity. The ball has a mass of $0.15$ kg at Earth’s surface.

Find the velocity $v(t)$ of the basevall at time $t$.
What is its velocity after $2$ seconds?

a. From the preceding discussion, the differential equation that applies in this situation is

$v′(t)=−g,$

where $g=9.8\, \text{m/s}^2$. The initial condition is $v(0)=v_0$, where $v_0=10$ m/s. Therefore the initial-value problem is $v′(t)=−9.8\,\text{m/s}^2,\,v(0)=10$ m/s.

The first step in solving this initial-value problem is to take the antiderivative of both sides of the differential equation. This gives

\[\int v′(t)\,dt=∫−9.8\,dt \nonumber \]

$v(t)=−9.8t+C.$

The next step is to solve for $C$. To do this, substitute $t=0$ and $v(0)=10$:

\[ \begin{align*} v(t) &=−9.8t+C \\[4pt] v(0) &=−9.8(0)+C \\[4pt] 10 &=C. \end{align*}\]

Therefore $C=10$ and the velocity function is given by $v(t)=−9.8t+10.$

b. To find the velocity after $2$ seconds, substitute $t=2$ into $v(t)$.

\[ \begin{align*} v(t)&=−9.8t+10 \\[4pt] v(2)&=−9.8(2)+10 \\[4pt] v(2) &=−9.6\end{align*}\]

The units of velocity are meters per second. Since the answer is negative, the object is falling at a speed of $9.6$ m/s.

Exercise $\PageIndex{6}$

Suppose a rock falls from rest from a height of $100$ meters and the only force acting on it is gravity. Find an equation for the velocity $v(t)$ as a function of time, measured in meters per second.

What is the initial velocity of the rock? Use this with the differential equation in Example $\PageIndex{6}$ to form an initial-value problem, then solve for $v(t)$.

$v(t)=−9.8t$

A natural question to ask after solving this type of problem is how high the object will be above Earth’s surface at a given point in time. Let $s(t)$ denote the height above Earth’s surface of the object, measured in meters. Because velocity is the derivative of position (in this case height), this assumption gives the equation $s′(t)=v(t)$. An initial value is necessary; in this case the initial height of the object works well. Let the initial height be given by the equation $s(0)=s_0$. Together these assumptions give the initial-value problem

\[ s′(t)=v(t),s(0)=s_0. \nonumber \]

If the velocity function is known, then it is possible to solve for the position function as well.

Example $\PageIndex{7}$: Height of a Moving Baseball

A baseball is thrown upward from a height of $3$ meters above Earth’s surface with an initial velocity of $10m/s$, and the only force acting on it is gravity. The ball has a mass of $0.15$ kilogram at Earth’s surface.

Find the position $s(t)$ of the baseball at time $t$.
What is its height after $2$ seconds?

We already know the velocity function for this problem is $v(t)=−9.8t+10$. The initial height of the baseball is $3$ meters, so $s_0=3$. Therefore the initial-value problem for this example is

To solve the initial-value problem, we first find the antiderivatives:

\[∫s′(t)\,dt=∫(−9.8t+10)\,dt \nonumber \]

$s(t)=−4.9t^2+10t+C.$

Next we substitute $t=0$ and solve for $C$:

$s(t)=−4.9t^2+10t+C$

$s(0)=−4.9(0)^2+10(0)+C$

Therefore the position function is $s(t)=−4.9t^2+10t+3.$

b. The height of the baseball after $2$ sec is given by $s(2):$

$s(2)=−4.9(2)^2+10(2)+3=−4.9(4)+23=3.4.$

Therefore the baseball is $3.4$ meters above Earth’s surface after $2$ seconds. It is worth noting that the mass of the ball cancelled out completely in the process of solving the problem.

Key Concepts

A differential equation is an equation involving a function $y=f(x)$ and one or more of its derivatives. A solution is a function $y=f(x)$ that satisfies the differential equation when $f$ and its derivatives are substituted into the equation.
A differential equation coupled with an initial value is called an initial-value problem. To solve an initial-value problem, first find the general solution to the differential equation, then determine the value of the constant. Initial-value problems have many applications in science and engineering.

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Differential Equations : Initial-Value Problems

Study concepts, example questions & explanations for differential equations, all differential equations resources, example questions, example question #1 : initial value problems.