Eigenvalues of an equation

[krocho]

hi I have the following eigenvalue problem
-(x²y')'=λy for 1<x<2
y(1)=y(2)=0


I tried plugging an equation y=x^a
and you get the equation
a²+a+λ=0
so for this I get that λ<1/4 to hava a solution. So does this mean, every λ smaller than 1/4 is an eigenvalue?
do you know what else I could do?

thanks

 

[HallsofIvy]

hi I have the following eigenvalue problem
-(x²y')'=λy for 1<x<2
y(1)=y(2)=0


I tried plugging an equation y=x^a
and you get the equation
a²+a+λ=0
so for this I get that λ<1/4 to hava a solution. So does this mean, every λ smaller than 1/4 is an eigenvalue?
do you know what else I could do?

thanks

Why should $$\lambda$$ be less than 1/4? That would make the powers of x real numbers but why would that be necessary? In fact, if the powers of x were real numbers wouldn't that make it impossible to satisfy y(1)= y(2)= 0?

What do solutions to such an equation look like if the characteristic equation has complex roots? Hint: the change of variable t= ln(x) converts an "Euler-type" equation to an equation with constant coefficients having the same characteristic equation.

Also, since this is a second order linear equation, it has exactly 2 eigenvalues.