MHB Eigenfunctions and Boundary Conditions for $$\frac{d^2}{dx^2}(xy) - λxy=0$$

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consider $$\frac{d^2}{dx^2}(xy) - λxy=0$$. Show eigenfunctions are $$y_{n}=\frac{\sin(n\pi x)}{x}$$. Boundary conditions are y(1)=0 and y regular at x=0

I integrated twice to obtain $$6xy=λx^3y+6Ax+6B$$ where A,B constants. I can't apply the condition y is regular because I don't know what it means pratically. Besides, I can't see how I can get the required solution from this equation.
 
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Your "twice integrating" is invalid, because you don't know what $y$ is, aside from being a function of $x$. I suggest letting a new function $u=u(x)=xy(x)$. Then the DE becomes $u''=\lambda u$. What are the solutions of that DE?

Can't help you with the "regular at $x=0$", other than it sounds like something to do with differentiability at the origin.
 
It says, as a boundary condition, y(x) must be regular at the singular point x=0. I see my mistake integrating, I was treating y as a constant w.r.t x.
 
I've got a line on the "regularity" thing. Solve the DE I posted in my previous post. What do you get?
 
ok so characteristic equation is m^2=λ, so 3 cases to consider. λ=0, λ<0 and λ>0. I can solve when I know boundary conditions.
 
You are not asked to solve the equation- you are given a possible solution and asked to show that it does, in fact, satisfy the differential equation. That requires that you differentiate the given function, not integrate anything.
 
You're wrong. It asks 'show that the eigenfuctions are ...' i.e you have to show there are no others.
 
Poirot said:
You're wrong. It asks 'show that the eigenfuctions are ...' i.e you have to show there are no others.

I could be wrong, but I don't think you have to show uniqueness. Eigenvectors, for example, are not unique. In fact, a scalar times an eigenvector is an eigenvector. Similarly, in this case, at the very least, a constant times an eigenfunction is an eigenfunction. Moreover, I think you show that the functions $e^{in\pi x}/x$ are eigenfunctions.
 
Following Ackbach's comment #2 above, let $u=xy$. Then $u''=\lambda u$. That is an SHM equation with solution $u=A\cos(\omega x) + B\sin(\omega x)$, where $\omega = \sqrt{-\lambda}$.

Thus $y = \dfrac{A\cos(\omega x) + B\sin(\omega x)}x$. The boundary condition that $y$ is regular at $x=0$ means that $y$ should not go to infinity at $x=0$. That tells you that $A=0$. The other boundary condition $y(1)=0$ then tells you that $\omega=n\pi$ for some integer $n$. Therefore the eigenfunctions are $\dfrac{\sin(n\pi x)}x$, with corresponding eigenvalues $-n^2$.
 
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