hi
I need your help to resolve this question
y''-y'-2y=18e^-t sin3t
y(c)=0 , y'(c)=3
?
I don't see how that rule applies to this problem. Since we don't know the intial contidtions at zero we assume
$\displaystyle y(0)=c_1 \quad y'(0)=c_2$
If we take the laplace transfrom of the equation we get
$\displaystyle s^2Y-sc_1-c_2-sY+c_1-2Y=\frac{54}{(s+1)^2+9}$
Solving for Y gives
$\displaystyle Y=\frac{c_2-c_1}{(s-2)(s+1)}+\frac{c_1s}{(s-2)(s+1)}+\frac{54}{(s-2)(s+1)[(s+1)^2+9]}$
Edit: You could use the theorem to invert the 2nd of the two terms in this problem.
First of all you have an arithmetic error. When you add 3 to the other side and make a common denominator, it should be $\displaystyle \displaystyle \begin{align*} \frac{3\left[ (s + 1)^2 + 9 \right]}{(s + 1)^2 + 9} \end{align*}$, which when you expand it out gives $\displaystyle \displaystyle \begin{align*} \frac{3(s + 1)^2 + 27}{(s + 1)^2 + 9} \end{align*}$. As for the rest, you need to solve for $\displaystyle \displaystyle \begin{align*} Y(s) \end{align*}$ then take the inverse transform to get back to $\displaystyle \displaystyle \begin{align*} y(t) \end{align*}$.
In the diff eq, change t to t+c and take LaplaceT and solve for y(t).
$\displaystyle \mathcal{L}_t[y(t)](s)$=$\displaystyle \frac{3 e^{-c} \left(e^c s^2+2 e^c s+6 s \sin (3 c)+10 e^c+6 \sin (3 c)+18 \cos (3 c)\right)}{\left(s^2-s-2\right) \left(s^2+2 s+10\right)}$
and an inverse LapT would look like:
$\displaystyle y(t)=e^{-3 c-t} \left(e^{c+3 t}+e^{3 t} \sin (3 c)-e^{3 c} \sin (3 t)+e^{3 t} \cos (3 c)+e^{3 c} \cos (3 t)-e^{4 c}-2 e^{3 c} \cos (3 c)\right)$