Differential equation, Laplace transform

[rayman123]

Homework Statement

solve the differential equation by using Laplace transform


$$x^{''}(t)+7x^{'}(t)+10x(t)=sin3t$$
$$x(0)=0, x^{'}(0)=0$$

Homework Equations

$$X(s)s^2+X(s)7s+X(s)10=\frac{3}{s^2+9}$$
then i get
$$X(s) = \frac{3}{s^2+9}\cdot\frac{1}{(s+5)(s+2)}$$

$$\frac{3}{(s^2+9)(s+5)(s+2)}=\frac{As+B}{s^2+9}+\frac{C}{s+5}+\frac{D}{s+2}\\<br /> \begin{cases}A+C+D=0\\7A+2C+B+5D=0\\10A+7B+9C+9D=0\\10B+18C+45D=3\end{cases}\Rightarrow \begin{cases}A=-\frac{21}{442}\\B=\frac{3}{442}\\C=-\frac{1}{34}\\D=\frac{1}{13}\end{cases}$$


$$x(t) = =-\frac{21}{442}cos3t+\frac{3}{442}sin3t-\frac{1}{34}e^{-2t}+\frac{1}{13}e^{-5t}$$
1) well i made a quck check (used other method to calculate A, B) to se if the are correct
and this is what i get
$$x=A\cos3t+B\sin3t$$
$$x'=-3A\sin3t+3B\cos3t$$
$$x''=-9A\cos3t-9B\sin3t$$

and from this i obtain
$$A=-\frac{21}{442}$$
$$B=-\frac{1}{442}$$

can someone please show me where did I make a mistake??

The Attempt at a Solution

 

[cronxeh]

$$X(s) = \frac{-3(7s-1)}{442(s^2+9)}\ + \frac{1}{13(s+2)}\ - \frac{1}{34(s+5)}\ }<br /> <br /> \begin{cases}A=-\frac{21}{442}\\B=\frac{3}{442}\\C=-\frac{1}{34}\\D=\frac{1}{13}\end{cases}<br /> correct$$

$$x(t) = \frac{1}{442}(sin(3t)-21cos(3t))\ + \frac{e^{-2t}}{13}\ - \frac{e^{-5t}}{34}\$$

 

[vela]

You forgot about the factor of 3 on the top of the Laplace transform of sine

$$\frac{B}{s^2+9} = \frac{3/442}{s^2+9} = \frac{1}{442}\left(\frac{3}{s^2+9}\right) \rightarrow \frac{1}{442} \sin 3t$$

As for the sign of that term, I'll let you track that down yourself.