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.
Best Answer
$$y^{\prime \prime}+\lambda y=0$$ 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 $c_1,c_2$.