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I recently found this answer to a similar problem I'm currently working on. The problem is the following...

Find the eigenvalues and eigenfunctions for

$y^{\prime \prime}+\lambda y=0$

with the boundary conditions

$y^{\prime} (0)=0$ , $y^{\prime} (1)=0$


In the answer linked above, the Ansatz for

  • $\lambda <0$ is $y(x)=C_1e^{\sqrt{\lambda} x}+C_2e^{-\sqrt{\lambda} x}$
  • $\lambda >0$ is $y(x)=C_1\cos(\sqrt{\lambda}x)+C_2\sin(\sqrt{\lambda}x)$

But I don't really get why. Has someone got some background information/explanation why $y(x)$ must be of this form?

The case for $\lambda =0$ is clear to me.

Thanks in advance.

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  • $\begingroup$ Try to solve the characteristic polynomial $r^2+\lambda =0$ For $\lambda > 0$ and $\lambda <0$. $\endgroup$
    – MtGlasser
    Jan 24 at 0:21
  • $\begingroup$ Sorry for the basic question but why exactly $r^2+\lambda$? It makes sense because I get the Ansatz for $y(x)$ above but is this always the characteristic polynomial I've to look for? $\endgroup$ Jan 24 at 0:24
  • $\begingroup$ When you solve a differential equation with constant coefficients yes you need the polynomial characteristiic. Maybe you should read the wiki page on differential linear equations. $\endgroup$
    – MtGlasser
    Jan 24 at 0:32
  • $\begingroup$ $y''+by'+cy=0$ has the characteristic polynomial as $r^2+br+c=0$ You find $r$ then the solution is $y=ke^{rx}$ Maybe you should read this en.wikipedia.org/wiki/Linear_differential_equation $\endgroup$
    – MtGlasser
    Jan 24 at 0:33
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$$y^{\prime \prime}+\lambda y=0$$ We suppose that the solution is on the form $y=e^{rx}$ then you get: $$r^2e^{rx}++\lambda e^{rx}=0$$ $$e^{rx}(r^2+\lambda)=0$$ $$\implies r^2+ \lambda=0$$ Solve the characteristic polynomial. You have three cases: $\lambda >0, \lambda =0, \lambda<0$. Then the solution is: $$ \lambda <0 \implies y=c_1e^{r_1x}+c_2e^{r_2x}$$ $$ \lambda >0 \implies y=c_1 \cos ({r_1x)}+c_2\sin (r_2x)$$ Where $r_1,r_2$ are solution of the polynomial characteristic. Apply the initial conditions you'll find the value of the constants $c_1,c_2$.

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  • $\begingroup$ Makes sense now, thanks a lot! $\endgroup$ Jan 24 at 0:41
  • $\begingroup$ You are welcome. The wiki page is also interesting. $\endgroup$
    – MtGlasser
    Jan 24 at 0:42

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