Solution of Second Order Differential Equation with non-constant coeffecient How do we solve the differential equation
$y''-2(\sin x)y'-(\cos x-\sin^2x)y=0$ IVP: $y\left(\dfrac{\pi}{2}\right)=0$ , $ y'\left(\dfrac{\pi}{2}\right)=1$ ?
Its neither constant coeffecient, nor $Cauchy-Euler$. I really don't know how to proceed. 
We need to find the value of $y(0)$
 A: Hint
The boundary conditions made me thinking about a significant contribution of $\cos(x)$ in the solution.
Try using $y=e^{-\cos(x)}u(x)$ and see how the differential equation simplifies.
This is the same idea as  what user170231 suggested.
A: Hint:
Let $u=\cos x$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=-(\sin x)\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(-(\sin x)\dfrac{dy}{du}\right)=-(\sin x)\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)-(\cos x)\dfrac{dy}{du}=-(\sin x)\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}-(\cos x)\dfrac{dy}{du}=-(\sin x)\dfrac{d^2y}{du^2}(-\sin x)-(\cos x)\dfrac{dy}{du}=(\sin^2x)\dfrac{d^2y}{du^2}-(\cos x)\dfrac{dy}{du}$
$\therefore(\sin^2x)\dfrac{d^2y}{du^2}-(\cos x)\dfrac{dy}{du}+2(\sin^2x)\dfrac{dy}{du}-(\cos x-\sin^2x)y=0$
$(1-\cos^2x)\dfrac{d^2y}{du^2}+(2-2\cos^2x-\cos x)\dfrac{dy}{du}-(\cos x+\cos^2x-1)y=0$
$(1-u^2)\dfrac{d^2y}{du^2}-(2u^2+u-2)\dfrac{dy}{du}-(u^2+u-1)y=0$
$y=e^{-u}$ is the trivial solution of the above ODE.
Then solve it by reduction of order.
