## 17Calculus - Solve Differential Equations Using Laplace Transforms

1st Order

2nd/Higher Order

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Calculus 1 Practice

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Practice Exams

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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.

### Laplace Transforms Table

Laplace Transforms

$$f(t)$$

$$\displaystyle{ F(s) }$$

Practice

Basic Functions

$$t^n, ~ n = 1, 2, 3, \ldots$$

$$\displaystyle{ \frac{n!}{s^{n+1}} }$$

$$e^{at}$$

$$\displaystyle{ \frac{1}{s-a} }$$

$$\sin(\alpha t)$$

$$\displaystyle{ \frac{\alpha}{s^2 + \alpha^2} }$$

$$\cos(at)$$

$$\displaystyle{ \frac{s}{s^2 + a^2} }$$

$$\sinh(at)$$

$$\displaystyle{ \frac{a}{s^2 - a^2} }$$

$$\cosh(at)$$

$$\displaystyle{ \frac{s}{s^2 - a^2} }$$

Special Functions

$$\delta(t)$$ unit impulse

$$1$$

$$\delta(t-\tau)$$ shifted unit impulse

$$e^{-\tau s}$$

$$u(t)$$ unit step

$$\displaystyle{ \frac{1}{s} }$$

$$u(t-\tau)$$ shifted unit step

$$\displaystyle{ \frac{1}{s} e^{-\tau s} }$$

Combined Functions

$$e^{at}\sin(\alpha t)$$

$$\displaystyle{ \frac{\alpha}{(s-a)^2 + \alpha^2} }$$

$$f(t)u(t-a)$$

$$\displaystyle{ e^{-sa} \mathcal{L}\{ f(t+a) \} }$$

$$t^n e^{at}, ~ n = 1, 2, 3, \ldots$$

$$\displaystyle{ \frac{n!}{(s-a)^{n+1}} \} }$$

Derivatives and Integrals

$$f'(t)$$

$$sF(s) - f(0)$$

$$f''(t)$$

$$s^2F(s) - sf(0) - f'(0)$$

$$\int_0^t{ f(v)~dv }$$

$$\displaystyle{ \frac{F(s)}{s} }$$

$$\displaystyle{ f^{(n)}(t) }$$

$$\displaystyle{ s^nF(s) - s^{n-1}f(0) - }$$ $$\displaystyle{ s^{n-2}f'(0) - . . . - f^{(n-1)}(0) }$$

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

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

See the Laplace Transforms Involving Derivatives and Integrals page for more discussion on these equations.

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

Solve $$y'+2y = \sin(3t), y(0) = 0$$

Problem Statement

Solve $$y'+2y = \sin(3t), y(0) = 0$$

Solution

### 3641 video

video by blackpenredpen

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Solve $$y''-4y = 3t^2, y(0)=2, y'(0) = 1$$

Problem Statement

Solve $$y''-4y = 3t^2, y(0)=2, y'(0) = 1$$

Solution

### 3642 video

video by blackpenredpen

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Solve $$y''-2y'-15y=0, y(0) = -2, y'(0)=1$$

Problem Statement

Solve $$y''-2y'-15y=0, y(0) = -2, y'(0)=1$$

Solution

### 3643 video

video by blackpenredpen

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Solve $$y''+16y = \cos(4t), y(0)=0, y'(0)=1$$

Problem Statement

Solve $$y''+16y = \cos(4t), y(0)=0, y'(0)=1$$

Solution

### 3644 video

video by blackpenredpen

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Solve $$y''+y = u(t-3), y(0)=0, y'(0) = 1$$

Problem Statement

Solve $$y''+y = u(t-3), y(0)=0, y'(0) = 1$$

Solution

### 3645 video

video by blackpenredpen

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Solve $$\displaystyle{ y''-4y+5y=0, y(0)=1, y'(0)=2 }$$

Problem Statement

Solve $$\displaystyle{ y''-4y+5y=0, y(0)=1, y'(0)=2 }$$

Solution

### 3646 video

video by blackpenredpen

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

Problem Statement

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

Solution

### 646 video

video by Dr Chris Tisdell

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

Problem Statement

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

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

Problem Statement

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

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

Solve $$y''+4y=q(t), t > 0$$; $$y(0)=y'(0)=0$$ 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

Solve $$y''+4y=q(t), t > 0$$; $$y(0)=y'(0)=0$$ 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

Solve $$y''+4y=q(t), t > 0$$; $$y(0)=y'(0)=0$$ 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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Solve $$y'' - 3y' + 2y = r(t)$$; $$y(0)=1, y'(0)=3$$ where $$r(t) = u(t-1) - u(t-2)$$

Problem Statement

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

Solve $$y'' - 3y' + 2y = r(t)$$; $$y(0)=1, y'(0)=3$$ 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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Solve $$4y'' + y = g(t);$$ $$y(0)=3, y'(0)=-7$$

Problem Statement

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

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

Problem Statement

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

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

Problem Statement

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

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

Problem Statement

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

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

Problem Statement

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

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

Problem Statement

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

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

### Trig Formulas

The Unit Circle

The Unit Circle [wikipedia] Basic Trig Identities

Set 1 - basic identities

$$\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }$$

$$\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }$$

$$\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }$$

$$\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }$$

Set 2 - squared identities

$$\sin^2t + \cos^2t = 1$$

$$1 + \tan^2t = \sec^2t$$

$$1 + \cot^2t = \csc^2t$$

Set 3 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle formulas

$$\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }$$

$$\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }$$

Trig Derivatives

 $$\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }$$ $$\displaystyle{ \frac{d[\cos(t)]}{dt} = -\sin(t) }$$ $$\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }$$ $$\displaystyle{ \frac{d[\cot(t)]}{dt} = -\csc^2(t) }$$ $$\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }$$ $$\displaystyle{ \frac{d[\csc(t)]}{dt} = -\csc(t)\cot(t) }$$

Inverse Trig Derivatives

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

Trig Integrals

 $$\int{\sin(x)~dx} = -\cos(x)+C$$ $$\int{\cos(x)~dx} = \sin(x)+C$$ $$\int{\tan(x)~dx} = -\ln\abs{\cos(x)}+C$$ $$\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C$$ $$\int{\sec(x)~dx} =$$ $$\ln\abs{\sec(x)+\tan(x)}+C$$ $$\int{\csc(x)~dx} =$$ $$-\ln\abs{\csc(x)+\cot(x)}+C$$

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