# Eigenfunction equation

#### ijustlost

I've been going round in circles with this problem for days:

Find the eigenvalues and associated normalised eigenfunctions of the operator L:

$$L_y = x^2 y'' + 2 xy' + \frac{y}{4}$$

Boundary conditions $$y(1)=y(e)=0$$

So what I've done:
substitute $$x = \exp(t)$$

Then $$L_y = \frac{d^2y}{dt^2} + \frac{dy}{dt} + \frac{y}{4}$$ where the differentials are now y wrt t

The eigenvalue equation is $$L_y = X_n y$$ where $$X_n$$ are my eigenvalues

Then I solve $$y'' + y' + (1/4 - X_n)y = 0$$

and get $$y = \exp (-1/2 t) (A \exp(+\sqrt(X_n)t) + B \exp(-\sqrt(X_n)t)$$

Applying the boundary problems I find B = - A

and that $$X_n == 0$$. Which doesn't seem right!

Am I on the right track here or have I missed the point totally!

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#### ijustlost

Can anyone help? Please say if i need to give more info on the problem

#### HallsofIvy

I've been going round in circles with this problem for days:

Find the eigenvalues and associated normalised eigenfunctions of the operator L:

$$L_y = x^2 y'' + 2 xy' + \frac{y}{4}$$

Boundary conditions $$y(1)=y(e)=0$$

So what I've done:
substitute $$x = \exp(t)$$

Then $$L_y = \frac{d^2y}{dt^2} + \frac{dy}{dt} + \frac{y}{4}$$ where the differentials are now y wrt t

The eigenvalue equation is $$L_y = X_n y$$ where $$X_n$$ are my eigenvalues

Then I solve $$y'' + y' + (1/4 - X_n)y = 0$$
And your boundary conditions are now y(0)= y(1)= 0.

and get $$y = \exp (-1/2 t) (A \exp(+\sqrt(X_n)t) + B \exp(-\sqrt(X_n)t)$$

Applying the boundary problems I find B = - A

and that $$X_n == 0$$. Which doesn't seem right!

Am I on the right track here or have I missed the point totally!
You're doing fine except that you are ASSUMING that Xn is not negative (since you are taking its square root). What happens if Xn is negative?

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