perturbation of differential equation I am struggling with the following problem, which on first site looks easy, but I can't see it. 
Given the DE:
$$\frac {d^2y}{dx^2} +y =\frac{\cos 2x}{a+ \epsilon y}$$
with initial conditions: $y(-\pi/4) =y(\pi/4) = 0$, $a>0$ and $|\epsilon| \ll1$
By using the scaling: $y=\alpha z$ this may written in the form: $$\frac {d^2z}{dx^2} +z =\frac{\cos 2x}{1+ \delta z}$$
How is this done? And can $\alpha$ be expressed in terms of $a$? and $\delta$ in terms of $a$ and $\epsilon$? 
Any help appreciated..

Let's substitute $y=\alpha z$ into your equation:
$$\alpha\left(\frac {d^2z}{dx^2} +z\right) =\frac{\cos 2x}{a+ \alpha\epsilon z}$$
$$\frac {d^2z}{dx^2} +z =\frac{\cos 2x}{\alpha a+ \alpha^2\epsilon z}$$
Let $\alpha = 1/a$. Then
$$\frac {d^2z}{dx^2} +z =\frac{\cos 2x}{1+ \frac{\epsilon}{a^2} z}$$
Introduce $\delta = \epsilon / a^2$. Finally:
$$\frac {d^2z}{dx^2} +z =\frac{\cos 2x}{1+ \delta z}$$
