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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..

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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:

$$\frac {d^2z}{dx^2} +z =\frac{\cos 2x}{1+ \delta z}$$

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    $\begingroup$ Now I see where I went wrong, I forgot the $\alpha$ in front of the $\alpha\left(\frac {d^2z}{dx^2} +z\right)$. Thanks! $\endgroup$ Mar 23 '18 at 21:20

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