How to Solve this Simultaneous DE? Using laplace transform

Question

How to solve this Simultaneous DE using laplace transform

$x''+ y''+3x = 15e^{-t}$
$y''-4x'+3y = 15\sin(2t)$

with initial values $x(0)=35$, $x'(0)= -48$ and $y(0)=27$?

Answer 1

Notice that:

\begin{align} x''+ y''+3x = 15e^{-t} & \Rightarrow_{\mathcal{L}} s^2 X - sx(0) - x'(0) + s^2 Y - sy(0) + y'(0) + 3X= 15 \frac{1}{s+1}\\ & \Rightarrow s^2 X - 35s + 48 +s^2 Y - 27s + a + 3X = \frac{15}{s+1}. \end{align}

and \begin{align} y''-4x'+3y = 15\sin(2t) & \Rightarrow_{\mathcal{L}} s^2 Y - sy(0) + y'(0) - 4(sX - x(0)) + 3Y = 15 \frac{2}{s^2 + 4}\\ & \Rightarrow s^2 Y - 27s + a - 4sX + 140 + 3Y = \frac{30}{s^2+4}. \end{align}

Here I posed $y'(0) = a$ since you didn't provided its value.

Now, you must solve the following system with respect to $X$ and $Y$:

$$\begin{cases} s^2 X - 35s + 48 +s^2 Y - 27s + a + 3X = \frac{15}{s+1}\\ s^2 Y - 27s + a - 4sX + 140 + 3Y = \frac{30}{s^2+4} \end{cases}$$

Finally, you must apply the inverse Laplace transform on both $X$ and $Y$ in order to obtain $x(t)$ and $y(t)$.

It is rather hard for me to solve this if you don't specify $y'(0)$.