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Differential equation

Posted November 11th, 2007 by Isoscel
in

Use the Laplace transform to solve the given intial-value problem.

$ y"+4y= sint U(t-2\pi), y(0)= 1 \:y'(0)=0 $

‹ Differential equation Linear partial differential eqn ›

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Answer

On November 11th, 2007 Structure says:

Let $ Y(s)=\mathcal{L}(y(t)(s) $
Then $ \mathcal{L}(y'(t)(s)=sY(s)-1 $
$ \mathcal{L}(y"(t)(s)=s^2Y(s)-s $
Equation becomes
$ s^2Y(s)-s+4Y(s)=e^{-2\pi s}\frac{1}{s^2+1} $
or $ Y(s)=\frac{s}{s^2+4}+e^{-2\pi s}\frac{1}{s^2+1}\frac{1}{s^2+4}=\frac{s}{s^2+4}+\frac{1}{3}(e^{-2\pi s}\frac{1}{s^2+1}-e^{-2\pi s}\frac{1}{s^2+4}) $

$ y(t) = \cos (2t)U(t)+\frac{1}{3}\sin (t) U(t-2\pi )- \frac{1}{6}\sin (2t) U(t-2\pi ) $

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