## 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 $t=t_0$. This means the system is at equilibrium, so $y(t_0)=0$, and the compression of the shock absorber is not changing, so $y'(t_0)=0$. 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 $y(t_0)=y_0$ and $y(t_1)=y_1$. 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: $y''+3y'-4y=0$, $y(0)=1$, $y'(0)=-9$.

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

$y(x)=c_1e^{-4x}+c_2e^x$.

$y^\prime=-4c_1e^{-4x}+c_2e^x$.

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

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

Then $c_1=1-c_2$. Substituting this expression into the second equation, we see that

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

So, $c_1=2$ and the solution to the initial-value problem is

$y(x)=2e^{-4x}-e^x$.

Solve the initial-value problem $y''-3y'-10y=0$, $y(0)=0$, $y'(0)=7$.

$y(x)=-e^{-2x}+e^{-5x}$.

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:

$y^{\prime\prime}+6y^\prime+13y=0, \ y(0)=0, \ y^\prime(0)=2$.

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

$y(x)=e^{-3x}(c_1\cos2x+c_2\sin2x)$.

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

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

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

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

$y=e^{-3x}\sin2x$

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

$y(x)=e^{x}(2\cos3x-\sin3x)$

Figure 2. Graph of $y(x)=e^{x}(2\cos3x-\sin3x)$.

## 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.)

$y^{\prime\prime}+2y^\prime+y=0, \ y(0)=1, \ y^\prime(0)=0$

Solve the initial-value problem and graph the solution. What is the position of the mass at time $t=2$ sec? How fast is the mass moving at time $t=1$ 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

$y(t)=c_1e^{-t}+c_2te^{-t}$.

$y^\prime(t)=-c_1e^{-t}+c_2(-te^{-t}+e^{-t})$.

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

\begin{aligned} c_1&=1 \\ -c_1+c_2&= 0 \end{aligned}.

Thus, $c_1=1$, $c_2=1$, and the solution to the initial value problem is

$y(t)=e^{-t}+te^{-t}$.

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

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

$y^\prime(t)=-e^{-t}+e^{-t}-te^{-t}=-te^{-t}$.

Then $y^\prime(1)=e^{-1}\approx-0.3679$. At time $t=1$, the mass is moving upward at $0.3679$ 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 $t=0.3$ sec? How fast is it moving at time $t=0.1$ sec? In what direction?

$y^{\prime\prime}+14y^\prime+49y=0, \ y(0)=0, \ y^\prime(0)=1$

$y(t)=te^{-7t}$

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

## Example: solving a boundary-value problem

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

• $y(0)=0$, $y\left(\frac{\pi}4\right)=0$
• $y(0)=1$, $y\left(\frac{\pi}8\right)=0$
• $y\left(\frac{\pi}8\right)=0$, $y\left(\frac{3\pi}8\right)=0$

$y(x)=c_1\cos{4t}+c_2\sin{4t}$

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

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.

## Exercise

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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## 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.

So, how do you solve an initial value problem?  First, write out the differential equation. Next, find the starting condition (initial value). Then, isolate the derivative in the equation. Now, take the antiderivative on both sides of the equation. Finally, use the starting condition (initial value) to find the constant in the antiderivative.

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.

In this article, we’ll talk about initial value problems and what they are.  We’ll also look at the steps you can take to solve them, along with some examples to show how it’s done in practice.

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.

• 1. Write out the equation – if the IVP is given as a word problem, you might have to translate into an equation.  Sometimes, the equation will be given.  You may need to introduce variables to write your equation.
• 2. Identify the starting condition – an IVP always has a starting condition (an initial value for the function).  Sometimes, the starting condition will be given.  Other times, you might have to use context to figure it out.
• 3. Isolate the derivative – before you can take an antiderivative (next step), you need to isolate the derivative function.  This is just like solving for any other variable in an equation.
• 4. Take the antiderivative – you will need to know the rules for taking antiderivatives of various types of functions.  This can get more complicated for advanced ordinary differential equations.
• 5. Use the starting condition – taking the antiderivative in the previous step will introduce an unknown constant.  We use the starting condition in our equation to find the value of the constant.

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.

After 3.2 hours, you are 220 miles south of Boston.  How far from Boston were you after 1.5 hours?

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

Step 1: write out the equation.

We are not given any variables, so we will need our own.  Let’s use S for the speed of the car, P for the position of the car, and t for the time (in hours).

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:

Step 2: identify the starting condition.

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).

Step 3: isolate the derivative.

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

Step 4: take the antiderivative.

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:

• ∫P’(t)dt = ∫60dt
• P(t) = 60t + K

where K is an unknown constant.

Step 5: use the starting condition.

Now, we use the starting condition to solve for the constant K.  Remember from Step 2 that our starting condition was P = 220 at t = 3.2.

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

• P(3.2) = 60(3.2) + K
• 220 = 192 + K
• 220 – 192 = K

So, the entire equation is P(t) = 60t + 28.  This tells us how many miles (south) the car is from Boston at t hours.

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).

At time t = 1.5 hours, the car is P(1.5) = 60(1.5) + 28 = 90 + 28 = 118 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.

The acceleration due to gravity is -32 feet per second.  What is the height of the ball at 2.5 seconds?

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).

The equation tells us the acceleration A of the ball at a given time t.  So, we have the equation:

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

Remember that acceleration is the derivative of velocity, meaning A(t) = V’(t).  So, we can rewrite the equation as:

• V’(t) = -32

We are given that the starting velocity of the ball is -20 feet per second.  This implies a velocity of V = -20 feet per second at a time t = 0 seconds.

As an ordered pair, we can write this as (0, -20).

Since the derivative V’(t) is already isolated in the equation V’(t) = -32, we don’t need to do anything here.

• ∫V’(t)dt = ∫-32dt
• V(t) = -32t + K

Now, we use the starting condition to solve for the constant K.  Remember from Step 2 that our starting condition was V = -20 at t = 0.

• P(t) = -32t + K
• P(0) = 60(0) + K
• -20 = 0 + K

So, the entire equation is V(t) = -32t – 20.  This tells us the velocity of the ball at t seconds.

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:

• V(t) = -32t – 20

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

• P’(t) = -32t – 20

We are given that the starting position of the ball is a height of 500 feet above ground.  This implies a position of P = 500 feet at a time t = 0 seconds.

As an ordered pair, we can write this as (0, 500).

Since the derivative P’(t) is already isolated in the equation P’(t) = -32t – 20, we don’t need to do anything here.

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:

• ∫P’(t)dt = ∫(-32t – 20)dt
• P(t) = -32t 2 /2 – 20t + K
• P(t) = -16t 2 – 20t + K

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.

• P(0) = -16(0) 2 – 20(0) + K
• 500 = 0 – 0 + K

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

The height of the ball at t = 2.5 seconds is given by:

• P(t) = -16t 2 – 20t + 500
• P(2.5) = -16(2.5) 2 – 20(2.5) + 500
• P(2.5) = -16(6.25) – 50 + 500
• P(2.5) = -100 + 450
• P(2.5) = 350

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

Now you know what an initial value problem is and how to solve one.  You also know the steps to take so that you have the right tools to solve the problem.

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

Don’t forget to subscribe to my YouTube channel & get updates on new math videos!

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## 8.3: Solution of Initial Value Problems

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• Page ID 30766

• William F. Trench
• Trinity University

## 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.

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