Fredholm's Alternative

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Homework Statement


This is not a book problem but rather just a question about how the book got here.
[tex]\frac{d^2u}{dx^2}+u=e^{x} \quad \mbox{with} \quad u(0)=0 \quad \mbox{and} \quad u(\pi)=0[/tex]
[tex] \Rightarrow \frac{d^2\phi}{dx^2}=-\lambda \phi \quad \mbox{with} \quad \phi(0)=0 \quad \mbox{and} \quad \phi(\pi)=0 [/tex]
[tex]\frac{d^2\phi}{dx^2} + (\lambda+1)\phi=0.[/tex]
Therefore, [tex]\lambda+1 = (\frac{n\pi}{L})^2=n^2,\,\,\quad n=1,2,3...[/tex]
The book says it is now clear that [tex]\lambda=0[/tex] but it doesn't make sense to me. Can someone please explain why [tex]\lambda=0[/tex] please?

Homework Equations


[tex]-\alpha_{n} \lambda_n=\frac{\int_a^b f(x)\phi_n(x)\,dx}{\int_a^b\phi_n^2 \rho \, dx}[/tex]
 
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