Find the solution of y"+2y'+3y=sin(t)+delta(t-3pi); y(0)=0, y'(0)=0. (I got Y(s)(s^2+3s+2)=1/(s^2+1)+e^(-3pi*s) and I know that you have to divide s^2+3s+2 to the other side but I don't know what will look like. Please help.)

Question

loser66 · Answer

nope, I don't help you, hehehe, so what?? 
$$Y(s)= \frac{1}{(s^2+1)(s^2+3s+2)}+\frac{e^{-3\pi s}}{s^2+3s+2}$$
 2 terms, --------1st-------- and -----------2nd term-------
1st term , use partial fraction decompose:
$$=\frac{1}{(s^2 +1)(s+2)(s+1)}= \frac{As+B}{s^2+1}+\frac{C}{s+1}+\frac{D}{s+2}$$
solve for A,B,C,D
$$(As+B)(s+1)(s+2)+C(s^2+1)(s+2)+D(s+1)(s^2+1)=1$$
when  s =-1$\rightarrow $ C = $\dfrac{1}{2}$
when s = -2 $\rightarrow$ D = $-\dfrac{1}{5}$
when s =0 $\rightarrow$ B= $-\dfrac{1}{5}$
when s = 1$\rightarrow$ A = 0
therefore, the first term is 
$$-\frac{1}{5}\frac{1}{s^2+1}+\frac{1}{2}\frac{1}{s+1}-\frac{1}{5}\frac{1}{s+2}$$
take inverse laplace transform of them to get the first term of solution
2nd term, do the same