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Module 4: Differential Equations
Initial-value problems, learning outcomes.
- Identify an initial-value problem
- Identify whether a given function is a solution to a differential equation or an initial-value problem
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 [latex]{y}^{\prime }=2x[/latex], then [latex]y\left(3\right)=7[/latex] is an initial value, and when taken together, these equations form an initial-value problem. The differential equation [latex]y\text{''}-3{y}^{\prime }+2y=4{e}^{x}[/latex] 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 [latex]y\left(0\right)=2[/latex] and [latex]{y}^{\prime }\left(0\right)=-1[/latex]. 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 [latex]t[/latex], which represents time. Thus, a value of [latex]t=0[/latex] represents the beginning of the problem.
Example: Verifying a Solution to an Initial-Value Problem
Verify that the function [latex]y=2{e}^{-2t}+{e}^{t}[/latex] is a solution to the initial-value problem
For a function to satisfy an initial-value problem, it must satisfy both the differential equation and the initial condition. To show that [latex]y[/latex] satisfies the differential equation, we start by calculating [latex]{y}^{\prime }[/latex]. This gives [latex]{y}^{\prime }=-4{e}^{-2t}+{e}^{t}[/latex]. Next we substitute both [latex]y[/latex] and [latex]{y}^{\prime }[/latex] into the left-hand side of the differential equation and simplify:
This is equal to the right-hand side of the differential equation, so [latex]y=2{e}^{-2t}+{e}^{t}[/latex] solves the differential equation. Next we calculate [latex]y\left(0\right)\text{:}[/latex]
This result verifies the initial value. Therefore the given function satisfies the initial-value problem.
Watch the following video to see the worked solution to Example: Verifying a Solution to an Initial-Value Problem
You can view the transcript for this segmented clip of “4.1.5” here (opens in new window) .
Verify that [latex]y=3{e}^{2t}+4\sin{t}[/latex] is a solution to the initial-value problem
First verify that [latex]y[/latex] solves the differential equation. Then check the initial value.
In the previous example, the initial-value problem consisted of two parts. The first part was the differential equation [latex]{y}^{\prime }+2y=3{e}^{x}[/latex], and the second part was the initial value [latex]y\left(0\right)=3[/latex]. 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 [latex]C[/latex]. The family of solutions to the differential equation in the example is given by [latex]y=2{e}^{-2t}+C{e}^{t}[/latex]. This family of solutions is shown in Figure 2, with the particular solution [latex]y=2{e}^{-2t}+{e}^{t}[/latex] 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 [latex]t[/latex] < 0 as [latex]t[/latex] goes to negative infinity. For C > 0, the function changes direction and increases in a gentle curve as [latex]t[/latex] goes to infinity. Larger values of C have a tighter curve closer to the [latex]y[/latex]-axis and at a higher y value. For C = 0, the function goes to 0 as [latex]t[/latex] goes to infinity. For C < 0, the function continues to decrease as [latex]t[/latex] goes to infinity.](https://s3-us-west-2.amazonaws.com/courses-images/wp-content/uploads/sites/4175/2019/04/11233914/CNX_Calc_Figure_08_01_002.jpg "express the solution of the following initial value problem in terms of an integral")
Figure 2. A family of solutions to the differential equation [latex]{y}^{\prime }+2y=3{e}^{t}[/latex]. The particular solution [latex]y=2{e}^{-2t}+{e}^{t}[/latex] is labeled.
Example: Solving an Initial-value Problem
Solve the following initial-value problem:
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
We are able to integrate both sides because the y term appears by itself. Notice that there are two integration constants: [latex]{C}_{1}[/latex] and [latex]{C}_{2}[/latex]. Solving the previous equation for [latex]y[/latex] gives
Because [latex]{C}_{1}[/latex] and [latex]{C}_{2}[/latex] are both constants, [latex]{C}_{2}-{C}_{1}[/latex] is also a constant. We can therefore define [latex]C={C}_{2}-{C}_{1}[/latex], which leads to the equation
Next we determine the value of [latex]C[/latex]. To do this, we substitute [latex]x=0[/latex] and [latex]y=5[/latex] into our aforementioned equation and solve for [latex]C\text{:}[/latex]
Now we substitute the value [latex]C=2[/latex] into our equation. The solution to the initial-value problem is [latex]y=3{e}^{x}+\frac{1}{3}{x}^{3}-4x+2[/latex].
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.
Solve the initial-value problem
First take the antiderivative of both sides of the differential equation. Then substitute [latex]x=0[/latex] and [latex]y=8[/latex] into the resulting equation and solve for [latex]C[/latex].
[latex]y=\frac{1}{3}{x}^{3}-2{x}^{2}+3x - 6{e}^{x}+14[/latex]
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 [latex]F=ma[/latex], where [latex]F[/latex] represents force, [latex]m[/latex] represents mass, and [latex]a[/latex] represents acceleration), to derive an equation that can be solved.
Figure 3. For a baseball falling in air, the only force acting on it is gravity (neglecting air resistance).
In Figure 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, [latex]g[/latex], is approximately [latex]9.8{\text{m/s}}^{2}[/latex]. We introduce a frame of reference, where Earth’s surface is at a height of 0 meters. Let [latex]v\left(t\right)[/latex] represent the velocity of the object in meters per second. If [latex]v\left(t\right)>0[/latex], the ball is rising, and if [latex]v\left(t\right)<0[/latex], the ball is falling (Figure 4).
Figure 4. Possible velocities for the rising/falling baseball.
Our goal is to solve for the velocity [latex]v\left(t\right)[/latex] at any time [latex]t[/latex]. To do this, we set up an initial-value problem. Suppose the mass of the ball is [latex]m[/latex], where [latex]m[/latex] 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 [latex]\left(F=ma\right)[/latex]. Acceleration is the derivative of velocity, so [latex]a\left(t\right)={v}^{\prime }\left(t\right)[/latex]. Therefore the force acting on the baseball is given by [latex]F=m{v}^{\prime }\left(t\right)[/latex]. 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 [latex]{F}_{g}=\text{-}mg[/latex], since this force acts in a downward direction. Therefore we obtain the equation [latex]F={F}_{g}[/latex], which becomes [latex]m{v}^{\prime }\left(t\right)=\text{-}mg[/latex]. Dividing both sides of the equation by [latex]m[/latex] gives the equation
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 [latex]t=0[/latex]. This is denoted by [latex]v\left(0\right)={v}_{0}[/latex].
Example: Velocity of a Moving Baseball
A baseball is thrown upward from a height of [latex]3[/latex] meters above Earth’s surface with an initial velocity of [latex]10\text{m/s}[/latex], and the only force acting on it is gravity. The ball has a mass of [latex]0.15\text{kg}[/latex] at Earth’s surface.
- Find the velocity [latex]v\left(t\right)[/latex] of the baseball at time [latex]t[/latex].
- What is its velocity after [latex]2[/latex] seconds?
where [latex]g=9.8{\text{m/s}}^{2}[/latex]. The initial condition is [latex]v\left(0\right)={v}_{0}[/latex], where [latex]{v}_{0}=10\text{m/s}\text{.}[/latex] Therefore the initial-value problem is [latex]{v}^{\prime }\left(t\right)=-9.8{\text{m/s}}^{2},v\left(0\right)=10\text{m/s}\text{.}[/latex] The first step in solving this initial-value problem is to take the antiderivative of both sides of the differential equation. This gives
The next step is to solve for [latex]C[/latex]. To do this, substitute [latex]t=0[/latex] and [latex]v\left(0\right)=10\text{:}[/latex]
Suppose a rock falls from rest from a height of [latex]100[/latex] meters and the only force acting on it is gravity. Find an equation for the velocity [latex]v\left(t\right)[/latex] 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 the example: Velocity of a Moving Baseball to form an initial-value problem, then solve for [latex]v\left(t\right)[/latex].
[latex]v\left(t\right)=-9.8t[/latex]
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 [latex]s\left(t\right)[/latex] 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 [latex]{s}^{\prime }\left(t\right)=v\left(t\right)[/latex]. 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 [latex]s\left(0\right)={s}_{0}[/latex]. Together these assumptions give the initial-value problem
If the velocity function is known, then it is possible to solve for the position function as well.
Example: Height of a Moving Baseball
A baseball is thrown upward from a height of [latex]3[/latex] meters above Earth’s surface with an initial velocity of [latex]10\text{m/s}[/latex], and the only force acting on it is gravity. The ball has a mass of [latex]0.15[/latex] kilogram at Earth’s surface.
- Find the position [latex]s\left(t\right)[/latex] of the baseball at time [latex]t[/latex].
- What is its height after [latex]2[/latex] seconds?
Next we substitute [latex]t=0[/latex] and solve for [latex]C\text{:}[/latex]
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Expert Answer. I am att …. Express the solution of the following initial value problem in terms of an integral. Explain your reasoning. y' cos x, y (1)9 Which of the following is the solution to the initial value problem? Choose the correct answer below. OB. y (cos t+9) dt A. y= cos t dt+9 O D. y O C. y cos t dt+ 1 (cos t+1) dt Explain your ...
Math. Calculus. Calculus questions and answers. Express the solution of the following initial value problem in terms of an integral. Explain your reasoning. y′=cos2x,y (3)=8 Which of the following is the solution to the initial value protlem? Choose the correcl answer below. A. y=∫8xcos2tdt+3 B. y=∫3xcos2tdt+8 C. y=∫3x (cos2t+8)dt D. y ...
Question: Express the solution of the following initial value problem in terms of an integral.Explain your reasoning. Which of the following is the solution to the initial value problem? Choose the correct answer below OD. y- cos? tat7 Explain your reasoning This can be proven true using the Fundamental Theorem of Calculus.
Calculus. Calculus questions and answers. Express the solution of the following initial value problem in terms of an integral. Explain your reasoning. y' = cos x, y (9) = 4 ? Which of the following is the solution to the initial value problem? Choose the correct answer below. X O A. y= »- S (cos?t+) dt 2 cost dt +4 ca on va X C. y= S (cos?t+9 ...
Question: X)5.4.67 Express the solution of the following initial value problem in terms of an integral. Explain your reasoning y' = sec2 x, y(9) = 5 Which of the following is the solution to the initial value problem? Choose the correct answer belovw D. y= | (see t + 9) dt Fxnlain vour reasoning Question is complete.
Expressing the solution of an initial value problem as an integral. Ask Question Asked 7 years ago. ... IS there a way to the express the solution to this IVP as an integral? ordinary-differential-equations; Share. ... Equivalent integral equation to initial value problem. 0.
is an example of an initial-value problem. Since the solutions of the differential equation are y = 2x3 +C y = 2 x 3 + C, to find a function y y that also satisfies the initial condition, we need to find C C such that y(1) = 2(1)3 +C =5 y ( 1) = 2 ( 1) 3 + C = 5. From this equation, we see that C = 3 C = 3, and we conclude that y= 2x3 +3 y = 2 ...
Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin ... Solve problems from Pre Algebra to Calculus step-by-step ... initial value problem. en. Related Symbolab blog posts. Practice Makes Perfect. Learning math takes practice, lots ...
Plugging v(t) v ( t) from the second formula (1) ( 1) into the first fromula one obtains. y(t) = y0 +y1t +∫t 0 ∫τ 0 (1 −y2(s))dsdτ . (2) (2) y ( t) = y 0 + y 1 t + ∫ 0 t ∫ 0 τ ( 1 − y 2 ( s)) d s d τ . An easy interchange of the order of integration shows that for any function s ↦ g(s) s ↦ g ( s) one has. ∫t 0 ∫τ 0 g(s ...
Specify an adaptive method: solve {y' (x) = -2 y, y (0)=1} from 0 to 10 using r k f algorithm. Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music….
That's just alpha over s squared plus alpha squared. Now, the next thing we want to do is we want to separate out the Laplace transform of Y terms, or the Y of s terms. Actually, even better, let's get rid of these initial conditions. y of 0, and y prime of 0 is 0, so this term is 0.
Accumulation problems are solved using definite integrals. Imagine we are given the following information: The temperature of a soup is increasing at a rate of r ( t) = 30 e − 0.3 t degrees Celsius per minute (where t is the time in minutes). At time t = 0 , the temperature of the soup is 23 degrees Celsius. And imagine we are asked to find ...
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This page titled 7.3: Solution of Initial Value Problems is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by William F. Trench via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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 ...
Express the solution of the given initial value problem in terms of a convolution integral. 4)/' + 4y'+17y = g(t) y(0)=0 y'(0) = 0 ... If we use the theorem of the convolution integral, we can take the inverse Laplace transform to solve for y. Thus, ... Express the solution for the following initial value problem in terms of a convolution ...
The Banach fixed point theorem is then invoked to show that there exists a unique fixed point, which is the solution of the initial value problem. An older proof of the Picard-Lindelöf theorem constructs a sequence of functions which converge to the solution of the integral equation, and thus, the solution of the initial value problem.
Express the solutions of the initial value and in terms of integrals. $$\frac{d y}{d x}=\sec x, \quad y(2)=3.$$ 06:08 Express the solutions of the initial value problems in terms of integrals.
Express the solution of the following initial value problem in terms of an integral. dy = 9+X dx 3 y(9) 3 Determine the solution to the initial value problem y = This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts.
Question: Express the solution of the following initial value problem in terms of an integral. Explain your reasoning. y' -tan x, y (5)-3 Which of the following is the solution to the initial value problem? Choose the correct answer below O A. y tan t dt + 3 08. y=J (tan t + 5) dt 3 C) D. y # | (tan t + 3) dt Explain your reasoning This can be ...
See Answer. Question: Express the solution of the following initial value problem in terms of an integral. Explain your reasoning y' =cos x, y (9) = 3 Which of the following is the solution to the initial value problem? Choose the correct answer below. y CosIdt9 on ya leostatu oc. v. contra (cost+3) dt Explain your reasoning the constant ...
See Answer. Question: Express the solution of the following initial value problem in terms of an integral. Explain your reasoning. y' -cos x, y2)-7 Which of the following is the solution to the initial value problem? Choose the correct answer below. OA. y (cos2 t+7) dt ов.-I (cos21+2) dt ycos t+2) dt O D. y cos2 t dt+2 Explain your reasoning.
Question: Express the solution of the following initial value problem in terms of an integral. Explain your reasoning. y' = sin? x, y(9) = 3 Which of the following is the solution to the initial value problem? Choose the correct answer below. х х O A. y = B. y= sin? t dt +9 3 (sin?t+9) at m = sin? t d ſsin ac x) х х O C. t dt + 3 OD. y= |(sin? t +3) dt 9