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I had a stuck in solving problem for a week now, so need help.

Please help!

the problem is as follows.

In a closed interval [itex]I=[0,\pi][/itex], the 2-times continuously differentiable function [itex]\phi(x)[/itex] and [itex]\psi(x)[/itex] meet the following conditions (they're ranged in [itex]\mathbb{R}[/itex]).

[itex]\psi ''(x)+\psi(x)=\phi ''(x)+\phi(x)=0[/itex]

Assume [itex]f(x)[/itex] be a continuous function defined in [itex]I[/itex], and let [itex]G(x,y), u(x)[/itex] as followings.

[tex]G(x,y)=\Bigg\{\begin{array}{l}\phi(x)\psi(y)\quad (0\leq y\leq x\leq \pi) \\ \psi(x)\phi(y)\quad (0\leq x\leq y\leq \pi) \end{array}

\\\\

u(x)=\int^\pi_0 G(x,y)f(y)dy[/tex]

Now my question is, what would be a proof for the equation: [itex]u''(x)+u(x)=Wf(x)[/itex] (for [itex]\exists W[/itex] is a constant).

I found u''(x)+u(x) constantly becomes 0 (for reason that [itex]u''(x)=\displaystyle{\int^\pi_0 \frac{\partial^2G(x,y)}{\partial{x}^2}f(y)dy} = -u(x)[/itex]).

Any hidden concept or my ignorance makes this so hard?

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# Need help solving functional analysis problem

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