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17Calculus - Solve Differential Equations Using Laplace Transforms

ODE Techniques

1st Order

2nd/Higher Order

Laplace Transforms

Applications

Additional Topics

Practice

Calculus 1 Practice

Calculus 2 Practice

Practice Exams

Tools

Calculus Tools

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ODE Techniques

1st Order

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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.

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

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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

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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

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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

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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

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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

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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

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

Final Answer

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

Problem Statement

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

Solution

648 video

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Final Answer

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

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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 \(\displaystyle{ q(t) = \left\{\begin{array}{crl} t & & 0 \leq t \leq \pi/2 \\ \pi/2 & & t > \pi/2 \end{array} \right. }\)

Final Answer

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

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

Final Answer

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

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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 \( r(t) = u(t-1) - u(t-2) \)

Final Answer

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

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

Final Answer

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

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

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Final Answer

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

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

2259 video

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Final Answer

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

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

Problem Statement

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

Final Answer

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

Problem Statement

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

Solution

2261 video

video by Michel vanBiezen

Final Answer

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

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You CAN Ace Differential Equations

Topics You Need To Understand For This Page

Related Topics and Links

external links you may find helpful

laplace transforms youtube playlist

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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Multi-Variable Calculus

Differential Equations

Precalculus

Engineering

Circuits

Semiconductors

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Practice Instructions

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.

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