I think Gerry's idea is a good one.
First, as for the proof of his equality, sure you know this one:
$$
\sin (A+B) = \sin A \cos B + \cos A \sin B \ .
$$
From which, you deduce
$$
\sin (A-B) = \sin A \cos B - \cos A \sin B \ .
$$
Adding both equalities, you get
$$
\sin A \cos B = \frac{1}{2} \left( \sin (A+B) + \sin (A-B) \right) \ .
$$
Which in your case gives that your differential equation is
$$
4y'' -y = \frac{1}{2} \sin \frac{3x}{2} + \frac{1}{2} \sin \frac{x}{2} \ .
$$
To what we can safely apply the Laplace transform, obtaining
$$
4{\cal L} [y''] - {\cal L}[y] = \frac{1}{2} {\cal L} \left[ \sin \frac{3x}{2} \right] +
\frac{1}{2} {\cal L} \left[ \sin \frac{x}{2} \right]
\ .
$$
That is
$$
4 (s^2 Y(s) -sy(0) - y'(0)) - Y(s) = \frac{1}{2} \frac{3/2}{s^2 + 9/4} + \frac{1}{2} \frac{1/2}{s^2 + 1/4} \ ,
$$
where $Y(s) = {\cal L}[y]$. Now put $a = y(0)$ and $b= y'(0)$ and solve this equation for $Y(s)$:
\begin{align}
Y(s) &= \frac{3}{16} \frac{1}{(s^2 + 9/4)(s-1/2)(s+1/2) } + \frac{1}{8} \frac{1}{(s^2 + 1/4)(s-1/2)(s+1/2) } \\
&{} \qquad
+ \frac{as+b}{(s-1/2)(s+1/2)} \ .
\end{align}
Finally, apply the inverse Laplace transform to both sides in order to obtain the general solution of your differential equation:
\begin{align}
y(x) &= \frac{3}{16} {\cal L}^{-1} \left[ \frac{1}{(s^2 + 9/4)(s-1/2)(s+1/2)} \right] \\
&{} \qquad + \frac{1}{8} {\cal L}^{-1} \left[ \frac{1}{(s^2 + 1/4)(s-1/2)(s+1/2) } \right] \\
&{} \qquad + {\cal L}^{-1} \left[ \frac{as+b}{(s-1/2)(s+1/2)} \right] \ .
\end{align}
Now is a matter of time and patiente to compute the right hand side. :-)
For instance, you could write
$$
\frac{as+b}{(s-1/2)(s+1/2)} = \frac{A}{s-1/2} + \frac{B}{s+1/2}
$$
for some constants $A$ and $B$, depending on $a$ and $b$. Hence
\begin{align}
{\cal L}^{-1} \left[ \frac{as+b}{(s-1/2)(s+1/2)} \right] &= A {\cal L}^{-1} \left[ \frac{1}{s-1/2} \right] + B {\cal L}^{-1} \left[ \frac{1}{s+1/2} \right] \\
&= A e^{x/2} + B e^{-x/2} \ .
\end{align}
In order to find out constants $A$ and $B$, you could do the following:
$$
as + b = A\left( s+ \frac{1}{2} \right) + B \left( s-\frac{1}{2} \right) = (A+B)s + \frac{A-B}{2} \ .
$$
Hence
$$
A = \dfrac{a+2b}{2} \qquad \text{and} \qquad B = \dfrac{a-2b}{2} \ .
$$
As for the remaining terms, you could write:
$$
\frac{1}{(s^2 + 9/4)(s-1/2)(s+1/2)} = A \frac{s}{s^2 + (3/2)^2} + B \frac{3/2}{s^2 + (3/2)^2} + C \frac{1}{s- 1/2} + D \frac{1}{s+1/2}
$$
and
$$
\frac{1}{(s^2 + 1/4)(s-1/2)(s+1/2)} = A' \frac{s}{s^2 + (1/2)^2} + B' \frac{1/2}{s^2 + (1/2)^2} + C' \frac{1}{s- 1/2} + D' \frac{1}{s+1/2}
$$
for some real constants (that is, not depending on $a$ and $b$ or anything else) $A,B,C,D$ and $A',B',C',D'$, which I let you the pleasure to compute :-) . Applying the inverse Laplace transform everywhere, you'll obtain:
$$
{\cal L}^{-1} \left[ \frac{1}{(s^2 + 9/4)(s-1/2)(s+1/2)} \right] = A\cos\frac{3x}{2} + B \sin\frac{3x}{2} + Ce^{x/2} + D e^{-x/2}
$$
and
$$
{\cal L}^{-1} \left[ \frac{1}{(s^2 + 1/4)(s-1/2)(s+1/2)} \right] = A'\cos\frac{x}{2} + B' \sin\frac{x}{2} + C'e^{x/2} + D' e^{-x/2}
$$
Some last thoughts. Laplace transform is great for different reasons. One of them is that you can trace back through your computations the origin of the terms in your final solution. For instance, if you differential equation had just been the homogeneous one
$$
4y'' -y = 0 \ ,
$$
then you could get rid of the two first addends in your general solution of the non-homogeneous one. That is, the solution, depending on the initial conditions $y(0)$ and $y'(0)$, would just have been:
$$
y(x) = \frac{y(0) +2 y'(0)}{2} e^{x/2} + \frac{y(0)-2y'(0)}{2} e^{-x/2}
$$