Answer
$y=c_1\cos \dfrac{x}{2}+c_2\sin \dfrac{x}{2}-\dfrac{1}{3} \cos x$
Work Step by Step
a) $y_c=c_1\cos \dfrac{x}{2}+c_2\sin \dfrac{x}{2}$
and $y_P=A \cos x+B \sin x$
The given differential equation becomes:
$-3A \cos x-3B \sin x=\cos x$
This gives: $A=\dfrac{-1}{3}; B=0$
Then, $y=y_c+y_p=c_1\cos \dfrac{x}{2}+c_2\sin \dfrac{x}{2}-\dfrac{1}{3} \cos x$
b) $y_c=c_1\cos \dfrac{x}{2}+c_2\sin \dfrac{x}{2}$
$y_1=\cos \dfrac{x}{2}; y_2=\sin \dfrac{x}{2}$
$u_1=-\cos \dfrac{x}{2}-\dfrac{2}{3}(\cos \dfrac{x}{2})^3$ and $u_2=\sin \dfrac{x}{2}-\dfrac{2}{3}(\sin \dfrac{x}{2})^3$
Now, $y_p=-\cos \dfrac{x}{2}-\dfrac{2}{3}(\cos \dfrac{x}{2})^3 \cos \dfrac{x}{2}+\sin \dfrac{x}{2}-\dfrac{2}{3}(\sin \dfrac{x}{2})^3\sin \dfrac{x}{2}=-\dfrac{1}{3} \cos x$
Then, $y=y_c+y_p=c_1\cos \dfrac{x}{2}+c_2\sin \dfrac{x}{2}-\dfrac{1}{3} \cos x$
