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The problem in question is

$(1+x)y''(x)+\frac12y'(x) + \lambda y(x)=0$

subject to $y(0)=y(3)=0$.

Thanks to a suggested change of variable ($y(x)=u(\sqrt{1+x})$), I've managed to find the general solution to the equation: $y(x)=c_1\cos(2\sqrt{\lambda}\sqrt{1+x})+c_ 2\sin(2\sqrt{\lambda}\sqrt{1+x})$ for $c_1, c_2 \in \mathbb{R}$.

However, the boundary values don't help much to find the problem's solution (is it even possible?), so I don't know how to proceed in finding the eigenvalues and eigenfunctions. Does anyone have any suggestion?

Edit: this sums up to finding $\lambda$ such that the system $Sc = 0$ below admits solutions other than $c=0$:

$$Sc=0\Leftrightarrow\begin{bmatrix}\cos(2\sqrt{\lambda}) & \sin(2\sqrt{\lambda}) \\ \cos(4\sqrt{\lambda}) & \sin(4\sqrt{\lambda})\end{bmatrix}\begin{bmatrix}c_1 \\ c_2\end{bmatrix}=\begin{bmatrix}0\\0\end{bmatrix}$$

Taking the determinant of $S$, we get $\cos(2\sqrt{\lambda})\sin(4\sqrt{\lambda}) - \cos(4\sqrt{\lambda})\sin(2\sqrt{\lambda})=\sin(2\sqrt{\lambda})$. If $Sc=0$ for $c\neq0$, then $\det(S)=0$, so we're pretty much done:

$$\sin(2\sqrt{\lambda})=0\Leftrightarrow\lambda=\frac{k^2\pi^2}4,\, k\in\mathbb{Z}.$$

After this, we can find that $c_1=0$ and $c_2$ can be anything, so the solution to this problem would be $y(x) = c_2\sin(k\pi\sqrt{1+x})$.

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  • $\begingroup$ What do you mean? $y(0)=0$ lets us get rid of the $\cos(k\pi\sqrt{1+x})$ term, because then $c_1 = 0$. $y(3)=0$ gives us $0=c_2\sin(2k\pi)$, which means $c_2$ can be anything. What am I not understanding correctly? $\endgroup$ – AstlyDichrar Jun 8 at 13:16
  • $\begingroup$ You are right, for the eigenvalues my phase difference is a multiple of $\pi$ and thus can be removed from the solution formula. $\endgroup$ – LutzL Jun 8 at 13:20
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You get by introducing a specific phase constant into the solution formula that the arguments of the sine and cosine are zero at $x=0$ and thus the solution satisfying the left boundary condition as $$ y(x)=c\sin(2\sqrtλ(\sqrt{1+x}-1)). $$ Then for the right boundary condition you need $$ 0=y(3)=c\sin(2\sqrtλ)\implies 2\sqrtλ=k\pi,~~k\in\Bbb N_{>0}. $$


Why the phase term: The function class $y(x)=c_1\cos(ϕ(x))+c_2\sin(ϕ(x))$ is equal by basic trigonometric identities to the class $y(x)=d_1\cos(ϕ(x)+A)+d_2\sin(ϕ(x)+A)$. Now set $A=-ϕ(0)$ to get that $y(0)=0$ implies $d_1=0$, so that $y(x)=d_2\sin(ϕ(x)-ϕ(0))$.

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  • $\begingroup$ I was able to find the eigenvalues in a different way, but this seems pretty elegant as well. I'll edit my post to include my partial solution. $\endgroup$ – AstlyDichrar Jun 7 at 23:47

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