Answer
a) $y=c_1\cos \dfrac{x}{2}+c_2\sin \dfrac{x}{2}-\dfrac{1}{3} \cos x$
b) $y=c_1\cos \dfrac{x}{2}+c_2\sin \dfrac{x}{2}-\dfrac{1}{3} \cos x$
Work Step by Step
(a) $y_c=c_1\cos \dfrac{x}{2}+c_2\sin \dfrac{x}{2}$
The particular solution is: $y_p=A \cos x+B \sin x$
Now, $-3A \cos x-3B \sin x=\cos x$
$\implies A=\dfrac{-1}{3}; B=0$
Then, we have $y=y_c+y_p$
or, $y=c_1\cos \dfrac{x}{2}+c_2\sin \dfrac{x}{2}-\dfrac{1}{3} \cos x$
b) $y_c=c_1\cos (\dfrac{x}{2})+c_2\sin (\dfrac{x}{2})$
The particular solution is: $y_p=A \cos x+B \sin x$
Here, we have $u_1=-\cos (\dfrac{x}{2})-\dfrac{2}{3}(\cos (\dfrac{x}{2})^3; u_2=\sin (\dfrac{x}{2}) -\dfrac{2}{3}(\sin \dfrac{x}{2})^3$
Now, $y_p=-\cos (\dfrac{x}{2})-\dfrac{2}{3}(\cos \dfrac{x}{2})^3 \cos (\dfrac{x}{2}) +\sin (\dfrac{x}{2}) -\dfrac{2}{3}(\sin \dfrac{x}{2})^3\sin \dfrac{x}{2}$
or, $y_p=-\dfrac{1}{3} \cos x$
Then, we get $y=y_c+y_p$
or, $y=c_1\cos \dfrac{x}{2}+c_2\sin \dfrac{x}{2}-\dfrac{1}{3} \cos x$
