Put the eigenvalue function in self-adjoint form

[GGGGc]

Homework Statement

The original equation is -(1-x^(2))y’’+xy’=ky. How to put it in self-adjoint form?
Also, if let x=cos(theta) how to put that form in d^(2)y/dx^(2)=-ky form?

Relevant Equations

-(1-x^(2))y’’+xy’=ky

Here’s my work:
The integrating factor I find is (x^(2)-1)^1/2. The self adjoint form I find is
-d/dx (((1-x^(2))^(3/2))*dy/dx))=k(x^(2)-1)^(1/2).
Am I right?

 

[pasmith]

The left hand side is of the form $$-u^2y'' - uu'y' = -u(uy')'$$ for some $u$.

 

[WWGD]

Homework Statement: The original equation is -(1-x^(2))y’’+xy’=ky. How to put it in self-adjoint form?
Also, if let x=cos(theta) how to put that form in d^(2)y/dx^(2)=-ky form?
Relevant Equations: -(1-x^(2))y’’+xy’=ky

Here’s my work:
The integrating factor I find is (x^(2)-1)^1/2. The self adjoint form I find is
-d/dx (((1-x^(2))^(3/2))*dy/dx))=k(x^(2)-1)^(1/2).
Am I right?

Please use ## ## tags and code to render Latex