You CAN Ace Differential Equations | |
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17calculus > differential equations > laplace transforms |
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Differential Equations Alpha List
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Solve Differential Equations Using Laplace Transforms |
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If you haven't seen Laplace Transforms before, the calculus 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 |
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\(\displaystyle{ \mathcal{L}\{y'\} = sY(s) - y(0) }\) |
\(\displaystyle{ \mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0) }\) |
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 | |
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 | |
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 | |
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Practice Problems |
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Instructions - - Unless otherwise instructed, solve these initial value problems using Laplace transforms. Give your answers in exact, completely factored form.
In the following problems, \(u(t)\) is the unit step function (Heaviside function).
Level A - Basic |
Practice A01 | |
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\(y'+y=u(t-1)\); \(y(0)=0\) | |
solution |
Practice A02 | |
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\(\displaystyle{y^{(4)}-y=0}\); \(y(0)=0\), \(y'(0)=1\), \(y''(0)=0\), \(y'''(0)=0\) | |
solution |
Level B - Intermediate |
Practice B01 | |
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\(y''+4y=q(t), t > 0 \); \(y(0)=y'(0)=0\) where \(\displaystyle{ q(t) = \left\{ \begin{array}{crl} t & & 0 \leq t \leq \pi/2 \\ \pi/2 & & t > \pi/2 \end{array} \right.}\) | |
answer |
solution |