17Calculus - Solve Differential Equations Using Laplace Transforms
17Calculus
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.
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Laplace Transforms Table
Laplace Transforms | |||
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\( f(t) \) |
\(\displaystyle{ F(s) }\) |
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Basic Functions | |||
\( t^n, ~ n = 1, 2, 3, \ldots \) |
\(\displaystyle{ \frac{n!}{s^{n+1}} }\) |
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\( e^{at} \) |
\(\displaystyle{ \frac{1}{s-a} }\) |
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\( \sin(\alpha t) \) |
\(\displaystyle{ \frac{\alpha}{s^2 + \alpha^2} }\) |
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\( \cos(at) \) |
\(\displaystyle{ \frac{s}{s^2 + a^2} }\) |
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\( \sinh(at) \) |
\(\displaystyle{ \frac{a}{s^2 - a^2} }\) |
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\( \cosh(at) \) |
\(\displaystyle{ \frac{s}{s^2 - a^2} }\) |
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Special Functions | |||
\( \delta(t) \) unit impulse |
\( 1 \) |
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\( \delta(t-\tau) \) shifted unit impulse |
\( e^{-\tau s} \) |
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\( u(t) \) unit step |
\(\displaystyle{ \frac{1}{s} }\) |
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\( u(t-\tau) \) shifted unit step |
\(\displaystyle{ \frac{1}{s} e^{-\tau s} }\) |
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Combined Functions | |||
\( e^{at}\sin(\alpha t) \) |
\(\displaystyle{ \frac{\alpha}{(s-a)^2 + \alpha^2} }\) |
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\( f(t)u(t-a) \) |
\(\displaystyle{ e^{-sa} \mathcal{L}\{ f(t+a) \} }\) |
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\( t^n e^{at}, ~ n = 1, 2, 3, \ldots \) |
\(\displaystyle{ \frac{n!}{(s-a)^{n+1}} \} }\) |
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Derivatives and Integrals | |||
\( f'(t) \) |
\( sF(s) - f(0) \) |
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\( f''(t) \) |
\( s^2F(s) - sf(0) - f'(0) \) |
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\(\displaystyle{ f^{(n)}(t) }\) |
\(\displaystyle{ s^nF(s) - s^{n-1}f(0) - }\) \(\displaystyle{ s^{n-2}f'(0) - . . . - f^{(n-1)}(0) }\) |
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\( \int_0^t{ f(v)~dv }\) |
\(\displaystyle{ \frac{F(s)}{s} }\) |
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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 |
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\(\displaystyle{ \mathcal{L}\{y'(t)\} = sY(s) - y(0) }\) |
\(\displaystyle{ \mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0) }\) |
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 |
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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 |
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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 |
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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
blackpenredpen - 3641 video solution
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
blackpenredpen - 3642 video solution
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
blackpenredpen - 3643 video solution
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
blackpenredpen - 3644 video 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
blackpenredpen - 3645 video 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
blackpenredpen - 3646 video solution
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
Dr Chris Tisdell - 646 video solution
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 |
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Solve \( y^{(4)}-y=0 \); \( y(0) = 0 \), \( y'(0) = 1 \), \( y''(0) = 0 \), \( y'''(0) = 0 \)
Final Answer |
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\(\displaystyle{ y(t) = \frac{1}{2}\left[ \sin(t) + \sinh(t) \right] }\)
Problem Statement
Solve \( y^{(4)}-y=0 \); \( y(0) = 0 \), \( y'(0) = 1 \), \( y''(0) = 0 \), \( y'''(0) = 0 \)
Solution
Dr Chris Tisdell - 648 video solution
video by Dr Chris Tisdell |
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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 |
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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 |
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\(\displaystyle{ y(t) = \frac{1}{4} \left[ t- \frac{1}{2} \sin(2t) \right] - \frac{u(t-\pi/2)}{4}\left[(t-\pi/2) - \frac{1}{2} \sin[2(t-\pi/2)] \right] }\)
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
Dr Chris Tisdell - 647 video solution
video by Dr Chris Tisdell |
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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 |
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Solve \( y'' - 3y' + 2y = r(t) \); \(y(0)=1, y'(0)=3\) where \( r(t) = u(t-1) - u(t-2) \)
Final Answer |
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\(\displaystyle{ y(t) = -e^t + 2e^{2t} + u(t-1)\left[ \frac{1}{2} - e^{t-1} + \frac{1}{2}e^{2(t-1)} \right] - u(t-2)\left[ \frac{1}{2} - e^{t-2} + \frac{1}{2}e^{2(t-2)} \right] }\)
Problem Statement
Solve \( y'' - 3y' + 2y = r(t) \); \(y(0)=1, y'(0)=3\) where \( r(t) = u(t-1) - u(t-2) \)
Solution
Dr Chris Tisdell - 649 video solution
video by Dr Chris Tisdell |
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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 |
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Solve \( 4y'' + y = g(t); \) \( y(0)=3, y'(0)=-7 \)
Final Answer |
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\( y(t) = 3\cos(t/2) - 14\sin(t/2) + (1/2)\int_0^t{\sin(t/2)g(t-\tau)~d\tau} \)
Problem Statement
Solve \( 4y'' + y = g(t); \) \( y(0)=3, y'(0)=-7 \)
Solution
Krista King Math - 2242 video solution
video by Krista King Math |
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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 |
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Solve \( y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0 \)
Final Answer |
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\(\displaystyle{ y(t) = \frac{F_0}{\omega_0^2-\omega^2} \left[ \sin(\omega t) - \frac{\omega}{\omega_0}\sin(\omega_0 t) \right] }\)
Problem Statement
Solve \( y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0 \)
Solution
Michel vanBiezen - 2259 video solution
video by Michel vanBiezen |
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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 |
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Solve \( y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0 \)
Final Answer |
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\( y(t) = e^{-3t}(-3+4e^t-e^{2t}) \)
Problem Statement
Solve \( y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0 \)
Solution
Michel vanBiezen - 2261 video solution
video by Michel vanBiezen |
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Final Answer
\( y(t) = e^{-3t}(-3+4e^t-e^{2t}) \)
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Really UNDERSTAND Differential Equations
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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.
