17Calculus - Solve Differential Equations Using Laplace Transforms
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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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 \(\mathcal{L}\{y(t)\} = Y(s)\) |
\(\displaystyle{ \mathcal{L}\{ f^{(n)}(t) \} = }\) \(\displaystyle{ s^nF(s) - s^{n-1}f(0) - }\) \(\displaystyle{ s^{n-2}f'(0) - . . . - f^{(n-1)}(0) }\) |
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
\( y' + y = u(t-1) \); \( y(0) = 0 \)
Problem Statement |
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\( y' + y = u(t-1) \); \( y(0) = 0 \)
Solution |
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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 |
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\( y^{(4)}-y=0 \); \( y(0) = 0 \), \( y'(0) = 1 \), \( y''(0) = 0 \), \( y'''(0) = 0 \)
Final Answer |
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\(\displaystyle{ y(t) = \frac{1}{2}\left[ \sin(t) + \sinh(t) \right] }\)
Problem Statement |
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\( y^{(4)}-y=0 \); \( y(0) = 0 \), \( y'(0) = 1 \), \( y''(0) = 0 \), \( y'''(0) = 0 \)
Solution |
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648 video
video by Dr Chris Tisdell |
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Final Answer |
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\(\displaystyle{ y(t) = \frac{1}{2}\left[ \sin(t) + \sinh(t) \right] }\) |
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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 |
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\(y''+4y=q(t), t > 0 \); \(y(0)=y'(0)=0\) 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 |
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\(y''+4y=q(t), t > 0 \); \(y(0)=y'(0)=0\) where \(\displaystyle{ q(t) = \left\{ \begin{array}{crl} t & & 0 \leq t \leq \pi/2 \\ \pi/2 & & t > \pi/2 \end{array} \right. }\)
Solution |
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647 video
video by Dr Chris Tisdell |
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Final Answer |
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\(\displaystyle{ y(t) = \frac{1}{4} \left[ t- \frac{1}{2} \sin(2t) \right] - \frac{u(t-\pi/2)}{4}\left[(t-\pi/2) - \frac{1}{2} \sin[2(t-\pi/2)] \right] }\) |
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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 |
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\( y'' - 3y' + 2y = r(t) \); \(y(0)=1, y'(0)=3\) 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 |
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\( y'' - 3y' + 2y = r(t) \); \(y(0)=1, y'(0)=3\) where \( r(t) = u(t-1) - u(t-2) \)
Solution |
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649 video
video by Dr Chris Tisdell |
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Final Answer |
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\(\displaystyle{ y(t) = -e^t + 2e^{2t} + u(t-1)\left[ \frac{1}{2} - e^{t-1} + \frac{1}{2}e^{2(t-1)} \right] - u(t-2)\left[ \frac{1}{2} - e^{t-2} + \frac{1}{2}e^{2(t-2)} \right] }\) |
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\( 4y'' + y = g(t); \) \( y(0)=3, y'(0)=-7 \)
Problem Statement |
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\( 4y'' + y = g(t); \) \( y(0)=3, y'(0)=-7 \)
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 |
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\( 4y'' + y = g(t); \) \( y(0)=3, y'(0)=-7 \)
Solution |
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2242 video
video by Krista King Math |
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Final Answer |
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\( y(t) = 3\cos(t/2) - 14\sin(t/2) + (1/2)\int_0^t{\sin(t/2)g(t-\tau)~d\tau} \) |
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\( y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0 \)
Problem Statement |
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\( y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0 \)
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 |
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\( y'' + \omega_0^2 y = F_0 \sin(\omega t); y’(0)=0, y(0)=0 \)
Solution |
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2259 video
video by Michel vanBiezen |
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Final Answer |
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\(\displaystyle{ y(t) = \frac{F_0}{\omega_0^2-\omega^2} \left[ \sin(\omega t) - \frac{\omega}{\omega_0}\sin(\omega_0 t) \right] }\) |
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\( y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0 \)
Problem Statement |
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\( y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0 \)
Final Answer |
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\( y(t) = e^{-3t}(-3+4e^t-e^{2t}) \)
Problem Statement |
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\( y’ + 3y + 2\int_0^ty(\tau)~d\tau = 2e^{-3t}; y(0)=0 \)
Solution |
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2261 video
video by Michel vanBiezen |
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Final Answer |
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\( y(t) = e^{-3t}(-3+4e^t-e^{2t}) \) |
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You CAN Ace Differential Equations
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Topics Listed Alphabetically
Single Variable Calculus |
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Multi-Variable Calculus |
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Differential Equations |
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Precalculus |
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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.
