## 17Calculus - Solve Differential Equations Using Laplace Transforms

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If you haven't seen Laplace Transforms before, the Laplace Transform section of 17calculus has a complete discussion of what they are and how to work with them.

Using Laplace Transforms to solve differential equation initial value problems is a great way to streamline solutions and, for forcing functions that are discontinuous, they are about the only way to solve them. The equations we use are

Laplace Transforms of Derivatives

$$\displaystyle{ \mathcal{L}\{y'(t)\} = sY(s) - y(0) }$$

$$\displaystyle{ \mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) }$$

where $$\mathcal{L}\{y(t)\} = Y(s)$$

$$\displaystyle{ \mathcal{L}\{ f^{(n)}(t) \} = }$$ $$\displaystyle{ s^nF(s) - s^{n-1}f(0) - }$$ $$\displaystyle{ s^{n-2}f'(0) - . . . - f^{(n-1)}(0) }$$

The idea is that we are given a differential equation which we first transform using the Laplace Transform, we solve it in the s-domain and then perform an inverse Laplace transform back to the t-domain to get our answer. Here is a very quick video giving these steps in a flow-chart.

### PatrickJMT - Laplace Transform flow-chart [1min-33secs]

video by PatrickJMT

This video clip explains in detail how to solve differential equations using Laplace Transforms including a quick calculation of the Laplace transform of the first derivative function above.

### Dr Chris Tisdell - solve differential equations using Laplace Transforms [28min-50secs]

video by Dr Chris Tisdell

This video expands on the one above taking you step-by-step through this process and, at the end of the video, gives you a big picture of how to do this. This is a great video.

### Dr Chris Tisdell - big picture [12min-45secs]

video by Dr Chris Tisdell

Practice

Unless otherwise instructed, solve these initial value problems using Laplace transforms. Give your answers in exact, completely factored form.
Notes
1. In these problems, $$u(t)$$ is the unit step function (Heaviside function).
2. Some of these problems may require convolution to solve.

Basic

$$y' + y = u(t-1)$$; $$y(0) = 0$$

Problem Statement

$$y' + y = u(t-1)$$; $$y(0) = 0$$

Solution

### 646 video

video by Dr Chris Tisdell

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$$y^{(4)}-y=0$$; $$y(0) = 0$$, $$y'(0) = 1$$, $$y''(0) = 0$$, $$y'''(0) = 0$$

Problem Statement

$$y^{(4)}-y=0$$; $$y(0) = 0$$, $$y'(0) = 1$$, $$y''(0) = 0$$, $$y'''(0) = 0$$

$$\displaystyle{ y(t) = \frac{1}{2}\left[ \sin(t) + \sinh(t) \right] }$$

Problem Statement

$$y^{(4)}-y=0$$; $$y(0) = 0$$, $$y'(0) = 1$$, $$y''(0) = 0$$, $$y'''(0) = 0$$

Solution

### 648 video

video by Dr Chris Tisdell

$$\displaystyle{ y(t) = \frac{1}{2}\left[ \sin(t) + \sinh(t) \right] }$$

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Intermediate

$$y''+4y=q(t), t > 0$$; $$y(0)=y'(0)=0$$ where $$\displaystyle{ q(t) = \left\{ \begin{array}{crl} t & & 0 \leq t \leq \pi/2 \\ \pi/2 & & t > \pi/2 \end{array} \right. }$$

Problem Statement

$$y''+4y=q(t), t > 0$$; $$y(0)=y'(0)=0$$ where $$\displaystyle{ q(t) = \left\{ \begin{array}{crl} t & & 0 \leq t \leq \pi/2 \\ \pi/2 & & t > \pi/2 \end{array} \right. }$$

$$\displaystyle{ y(t) = \frac{1}{4} \left[ t- \frac{1}{2} \sin(2t) \right] - \frac{u(t-\pi/2)}{4}\left[(t-\pi/2) - \frac{1}{2} \sin[2(t-\pi/2)] \right] }$$

Problem Statement

$$y''+4y=q(t), t > 0$$; $$y(0)=y'(0)=0$$ where $$\displaystyle{ q(t) = \left\{ \begin{array}{crl} t & & 0 \leq t \leq \pi/2 \\ \pi/2 & & t > \pi/2 \end{array} \right. }$$

Solution

### 647 video

video by Dr Chris Tisdell

$$\displaystyle{ y(t) = \frac{1}{4} \left[ t- \frac{1}{2} \sin(2t) \right] - \frac{u(t-\pi/2)}{4}\left[(t-\pi/2) - \frac{1}{2} \sin[2(t-\pi/2)] \right] }$$

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$$y'' - 3y' + 2y = r(t)$$; $$y(0)=1, y'(0)=3$$ where $$r(t) = u(t-1) - u(t-2)$$

Problem Statement

$$y'' - 3y' + 2y = r(t)$$; $$y(0)=1, y'(0)=3$$ where $$r(t) = u(t-1) - u(t-2)$$

$$\displaystyle{ y(t) = -e^t + 2e^{2t} + u(t-1)\left[ \frac{1}{2} - e^{t-1} + \frac{1}{2}e^{2(t-1)} \right] - u(t-2)\left[ \frac{1}{2} - e^{t-2} + \frac{1}{2}e^{2(t-2)} \right] }$$

Problem Statement

$$y'' - 3y' + 2y = r(t)$$; $$y(0)=1, y'(0)=3$$ where $$r(t) = u(t-1) - u(t-2)$$

Solution

### 649 video

video by Dr Chris Tisdell

$$\displaystyle{ y(t) = -e^t + 2e^{2t} + u(t-1)\left[ \frac{1}{2} - e^{t-1} + \frac{1}{2}e^{2(t-1)} \right] - u(t-2)\left[ \frac{1}{2} - e^{t-2} + \frac{1}{2}e^{2(t-2)} \right] }$$

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$$4y'' + y = g(t);$$ $$y(0)=3, y'(0)=-7$$

Problem Statement

$$4y'' + y = g(t);$$ $$y(0)=3, y'(0)=-7$$

$$y(t) = 3\cos(t/2) - 14\sin(t/2) + (1/2)\int_0^t{\sin(t/2)g(t-\tau)~d\tau}$$

Problem Statement

$$4y'' + y = g(t);$$ $$y(0)=3, y'(0)=-7$$

Solution

### 2242 video

video by Krista King Math

$$y(t) = 3\cos(t/2) - 14\sin(t/2) + (1/2)\int_0^t{\sin(t/2)g(t-\tau)~d\tau}$$

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$$y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0$$

Problem Statement

$$y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0$$

$$\displaystyle{ y(t) = \frac{F_0}{\omega_0^2-\omega^2} \left[ \sin(\omega t) - \frac{\omega}{\omega_0}\sin(\omega_0 t) \right] }$$

Problem Statement

$$y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0$$

Solution

### 2259 video

video by Michel vanBiezen

$$\displaystyle{ y(t) = \frac{F_0}{\omega_0^2-\omega^2} \left[ \sin(\omega t) - \frac{\omega}{\omega_0}\sin(\omega_0 t) \right] }$$

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$$y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0$$

Problem Statement

$$y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0$$

$$y(t) = e^{-3t}(-3+4e^t-e^{2t})$$

Problem Statement

$$y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0$$

Solution

### 2261 video

video by Michel vanBiezen

$$y(t) = e^{-3t}(-3+4e^t-e^{2t})$$

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You CAN Ace Differential Equations

 basics of differential equations basics of laplace transforms convolution integral

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