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I am given this ODE:

$$\sin(y''+\epsilon y)+y=1+ \sin\epsilon +\epsilon \sin t$$

And it is given that:

$$y(0)= \cos \epsilon$$ $$y'(0)=\sin \epsilon$$

where $\epsilon \approx 0$ and $t\in[-\delta, \delta]$

It is clear this ODE should be linearized, but with the additional parameter and the inexplicit $y''$, I'm having trouble understanding how to tackle the problem correctly.

Any help would be greatly appreciated!

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Where did this come from? Regards –  Amzoti Feb 3 '13 at 3:04

1 Answer 1

Taking the initial conditions into account, we see that at $t=0$ $$\sin (y''+\epsilon y)=1-\cos \epsilon+\sin\epsilon$$ which is a small number. Therefore, $y''(0)+\epsilon y(0) \approx \pi n$ for some integer $n$, which we cannot determine from the information given. The linearization depends on $n$: it is $\sin (y''+\epsilon y) \approx (-1)^n (y''+\epsilon y - \pi n)$. Hence, the linearized equation is $$ y''+(\epsilon +(-1)^n) y = \pi n+ (-1)^n (1+\sin\epsilon +\epsilon \sin t) $$ The form of the solutions of the linearized equation is rather different for even $n$ and for odd $n$.

But maybe you should assume that $y''$ is small near $0$, which corresponds to taking $n=0$ above.

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