1. ## Separation of Variables

This is supposed to be done with separation of variables.

$\displaystyle x^2u_{xx}+y^2u_{yy}+5xu_x-5yu_y+4u=0$

$\displaystyle u(x,y)=\varphi(x)\psi(y)$

$\displaystyle x^2\varphi''(x)\psi(y)+y^2\varphi(x)\psi''(y)+5x\v arphi'(x)\psi(y)-5y\varphi(x)\psi'(y)+4\varphi(x)\psi(y)=0$

$\displaystyle \displaystyle\frac{x^2\varphi''(x)+5x\varphi'(x)}{ \varphi(x)}=-\left[\frac{y^2\psi''(y)-5y\psi'(y)+4\psi(y)}{\psi(y)}\right]$

I am not sure if this getting anywhere. Is there another way to tackle this or is this correct? If so, what next?

2. Originally Posted by dwsmith
This is supposed to be done with separation of variables.

$\displaystyle x^2u_{xx}+y^2u_{yy}+5xu_x-5yu_y+4u=0$

$\displaystyle u(x,y)=\varphi(x)\psi(y)$

$\displaystyle x^2\varphi''(x)\psi(y)+y^2\varphi(x)\psi''(y)+5x\v arphi'(x)\psi(y)-5y\varphi(x)\psi'(y)+4\varphi(x)\psi(y)=0$

$\displaystyle \displaystyle\frac{x^2\varphi''(x)+5x\varphi'(x)}{ \varphi(x)}=-\left[\frac{y^2\psi''(y)-5y\psi'(y)+4\psi(y)}{\psi(y)}\right]$

I am not sure if this getting anywhere. Is there another way to tackle this or is this correct? If so, what next?
Everything is perfect so far.

Now observe that you are equating two expressions that are defined in different variables. Thus, the only time they are equal is if they're equal to the same constant, call it $\displaystyle \lambda$.

Now, you can rewrite this as a pair of ODEs:

$\displaystyle x^2\varphi^{\prime\prime}(x)+5x\varphi^{\prime}(x) =\lambda\varphi(x) \implies x^2\varphi^{\prime\prime}(x)+5x\varphi^{\prime}(x)-\lambda\varphi(x)=0$

$\displaystyle y^2\psi^{\prime\prime}(y)-5y\psi^{\prime}(y)+4\psi(y)=-\lambda\psi(y) \implies y^2\psi^{\prime\prime}(y)-5y\psi^{\prime}(y)+\left(4+\lambda\right)\psi(y)=0$

Each of these ODEs are Cauchy-Euler equations. Do you know how to proceed?

3. Yes.

4. Originally Posted by Chris L T521
Everything is perfect so far.

Now observe that you are equating two expressions that are defined in different variables. Thus, the only time they are equal is if they're equal to the same constant, call it $\displaystyle \lambda$.

Now, you can rewrite this as a pair of ODEs:

$\displaystyle x^2\varphi^{\prime\prime}(x)+5x\varphi^{\prime}(x) =\lambda\varphi(x) \implies x^2\varphi^{\prime\prime}(x)+5x\varphi^{\prime}(x)-\lambda\varphi(x)=0$

$\displaystyle y^2\psi^{\prime\prime}(y)-5y\psi^{\prime}(y)+4\psi(y)=-\lambda\psi(y) \implies y^2\psi^{\prime\prime}(y)-5y\psi^{\prime}(y)+\left(4+\lambda\right)\psi(y)=0$

Each of these ODEs are Cauchy-Euler equations. Do you know how to proceed?
I like it, nice!