• May 8th 2012, 05:13 AM
cristi92
$y''-4y'+5y=sin(t), t \in [0,+\infty), y(0)=0, y'(0)=0$
$s^{2}Y(s)-sy(0)-y'(0)-4(sY(s)-y(0))+5\frac{1}{s^{2}}=\frac{1}{s^{2}+1} <=>$
$Y(s)(s^{2}-4s)=\frac{1}{s^{2}+1}-\frac{5}{s^{2}} <=>$
$Y(s)(s^{2}-4s)=\frac{-4s^{2}-5}{s^{2}(s^{2}+1)} <=>$
$Y(s)=\frac{-4s^{2}-5}{s^{2}(s^{2}+1)(s^{2}-4s)} <=>$
$Y(s)=\frac{-1}{5(s^{2}+1)}+\frac{5}{4s^{2}}+\frac{21}{80(s+2)}-\frac{21}{80(s-2)} <=>$
$y(t)=\frac{-1}{5}sin(t)+\frac{5}{4}t+\frac{21}{80}e^{-2t}-\frac{21}{80}e^{2t}$

Y(s) is the laplace transform of y(t) function.
Can you please tell me what I did wrong? The answer should be y''-4y'+5y=sin(t), y(0)=0, y'(0)=0 - Wolfram|Alpha .

EDIT:
$Y(s)=\frac{1}{29}\left ( \frac{-4}{s}+\frac{1}{s^{2}}+\frac{4}{s^{3}}+\frac{s-1}{s^2+1}-\frac{129}{s-4} \right )$ which means
$y(t)=\frac{1}{29}\left ( -4+t+2t^{2}+L^{-1}\left \{ \frac{s-1}{s^2+1} \right \} -129e^{4t} \right ) <=>$
$y(t)=\frac{1}{29}\left ( -4+t+2t^{2}+cos(t)-sin(t)-129e^{4t} \right )$

but the answer is still wrong...
• May 8th 2012, 07:07 AM
BobP
Mistake on the second line. The $LT$ of $5y$ is $5Y(s).$