Separation of Variables

**dwsmith** · January 4th 2011, 08:28 PM

This is supposed to be done with separation of variables.

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

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

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

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

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

**Chris L T521** · January 4th 2011, 08:42 PM

Originally Posted by dwsmith

This is supposed to be done with separation of variables.

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

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

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

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

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 $\lambda$ .

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

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

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

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

**dwsmith** · January 4th 2011, 08:46 PM

Yes.

**AllanCuz** · January 6th 2011, 09:33 AM

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 $\lambda$ .

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

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

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

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

I like it, nice!

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