Solve system differential equation using Laplace transform Question : 
Solve system differential equation using Laplace transform : 
$\cases{x'-3x+2y= \sin t\\4x-y'-y=\cos t\\x(0)=0,y(0)=0}$
My try : 
Take Laplace  we find : 
$\cases{(s-3)X(s)=\frac{1}{s^2+1}-2Y(s)\\(s+1)Y(s)=4X(s)-\frac{s}{1+s^2}}$
What's the fast way to solving this system : 
My work :
$Y(s)=4\frac{X(s)}{1+s}-\frac{s}{(1+s)(1+s^2)}$
we find : $X(s)=\frac{1-s}{(1+s^2)(s^2-2s+5)}$
Please If any one have ideas to solving above system tell because this killed my time in exams 
 A: Well, we have the following coupled DE's:
$$
\begin{cases}
\text{x}'\left(t\right)-3\cdot\text{x}\left(t\right)+2\cdot\text{y}\left(t\right)=\sin\left(t\right)\\
\\
4\cdot\text{x}\left(t\right)-\text{y}'\left(t\right)-\text{y}\left(t\right)=\cos\left(t\right)\\
\\
\text{x}\left(0\right)=0\\
\\
\text{y}\left(0\right)=0
\end{cases}\tag1
$$
Using Laplace transform, and the table of selected Laplace transforms we can transform equation $(1)$ to the s-domain:
$$
\begin{cases}
\text{s}\cdot\text{X}\left(\text{s}\right)-\text{x}\left(0\right)-3\cdot\text{X}\left(\text{s}\right)+2\cdot\text{Y}\left(\text{s}\right)=\frac{1}{1+\text{s}^2}\\
\\
4\cdot\text{X}\left(\text{s}\right)-\text{s}\cdot\text{Y}\left(\text{s}\right)-\text{y}\left(0\right)-\text{Y}\left(\text{s}\right)=\frac{\text{s}}{1+\text{s}^2}\\
\\
\text{x}\left(0\right)=0\\
\\
\text{y}\left(0\right)=0
\end{cases}\tag2
$$
Using the initial conditions, we get:
$$
\begin{cases}
\text{s}\cdot\text{X}\left(\text{s}\right)-3\cdot\text{X}\left(\text{s}\right)+2\cdot\text{Y}\left(\text{s}\right)=\frac{1}{1+\text{s}^2}\\
\\
4\cdot\text{X}\left(\text{s}\right)-\text{s}\cdot\text{Y}\left(\text{s}\right)-\text{Y}\left(\text{s}\right)=\frac{\text{s}}{1+\text{s}^2}\\
\end{cases}\tag3
$$
Solving for $\text{X}\left(\text{s}\right)$ and $\text{Y}\left(\text{s}\right)$, we get:
$$
\begin{cases}
\text{X}\left(\text{s}\right)=\frac{\frac{1}{1+\text{s}^2}-2\text{Y}\left(\text{s}\right)}{\text{s}-3}\\
\\
\text{Y}\left(\text{s}\right)=-\frac{\frac{\text{s}}{1+\text{s}^2}-4\text{X}\left(\text{s}\right)}{\text{s}+1}\\
\end{cases}\tag4
$$
Subsituting them into each other gives:


*

*$$\text{X}\left(\text{s}\right)=\frac{\frac{1}{1+\text{s}^2}-2\left(-\frac{\frac{\text{s}}{1+\text{s}^2}-4\text{X}\left(\text{s}\right)}{\text{s}+1}\right)}{\text{s}-3}\tag5$$

*$$\text{Y}\left(\text{s}\right)=-\frac{\frac{\text{s}}{1+\text{s}^2}-4\left(\frac{\frac{1}{1+\text{s}^2}-2\text{Y}\left(\text{s}\right)}{\text{s}-3}\right)}{\text{s}+1}\tag6$$

Steps you need to take: solve both the above equations for the function that your looking for and perform the inverse Laplace transform on them.

A: First, we note that $X(s)=\frac{1-s}{(1+s^2)(s^2-2s+5)}$
has partial fraction decomposition:
$$
\frac{s-5}{10\left( {{s}^{2}}-2s+5\right) }-\frac{s-3}{10\left( {{s}^{2}}+1\right) }.
$$
Next, simply look up in the Inverse Laplace transform table.
