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Differential equation

Posted November 12th, 2007 by Isoscel
in

Use the Laplace transform to solve the given initial value problem

65. y'+2y=f(t) y(0)=0 f(t)= {t, 0 < t< 1; 0 t > 1}
Solution
$ f(t)=tu(t)-tu(t-1) $
$ \mathcal{L}(y(t))(s)=Y(s) $
$ \mathcal{L}(y'(t))(s)=sY(s)-y(0)=sY(s) $
$ \mathcal{L}(f(t))(s)=\frac{1}{s^2}-(\frac{1}{s}+\frac{1}{s^2})e^{-s} $
$ sY(s)+2Y(s)=\frac{1}{s^2}-(\frac{1}{s}+\frac{1}{s^2})e^{-s} $
$ Y(s)=\frac{1}{s^2(s+2)}-(\frac{1}{s(s+2)}+\frac{1}{s^2(s+2)})e^{-s} $
$ Y(s)=\frac{1}{2s^2}-\frac{1}{4s}+\frac{1}{4(s+2)}+(-\frac{1}{2s^2}-\frac{1}{4s}+\frac{1}{4(s+2)})e^{-s} $
$ y(t)=(\frac{t}{2}-\frac{1}{4}+\frac{1}{4}e^{-2t})u(t)-(\frac{t-1}{2}+\frac{1}{4}-\frac{1}{4}e^{-2(t-1)})u(t-1) $

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