<s>[INST] How to Solve this Simultaneous DE? Using laplace transform 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 $y(0)=27$?
 [/INST] 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 $y'(0)$.
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