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[tex]

\frac{d^{2}u}{dt^{2}}=\lambda u, 0<x<L, \frac{du}{dx}(0) = 0, u(0)=L

[/tex]

**[tex]\lambda > 0[/tex]**

[tex]

u = c_{1}cosh(\sqrt{\lambda}x)+c_{2}sinh(\sqrt{\lambda}x)

[/tex]

[tex]

\frac{du}{dx} = c_{1}\sqrt{\lambda}sinh(\sqrt{\lambda}x)+c_{2}\sqrt{\lambda}cosh(\sqrt{\lambda}x)

[/tex]

[tex]

\frac{du}{dx}(0) = c_{2}=0;

u(0)=u(L)\Rightarrow c_{1}cosh(\sqrt{\lambda}x)=c_{1}

[/tex]

It appears I cant find c1 unless i did something wrong...

**[tex] \lambda = 0[/tex]**

[tex]u = c_{1}+c_{2}x[/tex]

[tex]\frac{du}{dx}(0) = 0\Rightarrow c_{1}=0[/tex]

[tex]u(L) = u(0) \Rightarrow c_{2}=0[/tex]

**[tex]\lambda < 0[/tex]**

[tex] u = c_{1}cos(\sqrt{\lambda}x)+c_{2}sin(\sqrt{\lambda}x)[/tex]

[tex] \frac{du}{dx}(0) = \sqrt{lambda}c_{2}cos(\sqrt{\lambda}0) \Rightarrow c_{2}=0[/tex]

[tex] u(0) = u(L) \Rightarrow c_{1}=c_{1}cos(\sqrt{\lambda}L)[/tex]

[tex] \lambda = \left(\frac{arccos(1)}{L}\right)^{2} [/tex]

[tex]u = cos\left(\frac{arccos(1)x}{L}\right)[/tex]

is that right? i've never seen an answer like that... Any help is appreciated!