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) $y_c=c_1\cos \dfrac{x}{2}+c_2\sin \dfrac{x}{2}\\ y_P=A \cos x+B \sin x$
Write the given differential equation becomes:
$-3A \cos x-3B \sin x=\cos x$
This gives: $A=\dfrac{-1}{3}; B=0$
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) $y_c=c_1\cos \dfrac{x}{2}+c_2\sin \dfrac{x}{2}$
This gives: $y_1=\cos \dfrac{x}{2}; y_2=\sin \dfrac{x}{2}$
$u_1=-\cos \dfrac{x}{2}-\dfrac{2}{3} \times (\cos \dfrac{x}{2})^3\\ u_2=\sin \dfrac{x}{2}-\dfrac{2}{3} \times (\sin \dfrac{x}{2})^3$
Now, $y_p=-\cos [\dfrac{x}{2}] -\dfrac{2}{3}(\cos \dfrac{x}{2})^3 \cos [\dfrac{x}{2}]+\sin \dfrac{x}{2}-\dfrac{2}{3}(\sin \dfrac{x}{2})^3\sin \dfrac{x}{2}=-\dfrac{1}{3} \times \cos x$
Then, we have $y=y_c+y_p$
or, $y=c_1\cos \dfrac{x}{2}+c_2\sin \dfrac{x}{2}-\dfrac{1}{3} \cos x$
