# Partial differential equation

• Nov 22nd 2010, 08:49 PM
mukmar
Partial differential equation
I'm trying to remember how to solve a type of equation I've reduced a second order partial differential to.

In the notes it goes from the equation:
$\displaystyle 2\xi u_{\xi\eta} - u_{\eta} = 0$

to
$\displaystyle 2\xi u_{\xi\eta} - u_{\eta} = 2\xi^{3/2}\frac{d}{d\xi}\left(\frac{u_{\eta}}{\xi&{1/2}}\right) = 0$

With subscripts denoting partial derivatives.

I know that these two steps are equivalent, I'm just wondering on what the procedure used is to get the second equivalent form.

If the procedure has a name which I can look up.

Any assistance would be greatly appreciated, thank you.
• Nov 23rd 2010, 02:06 AM
BobP
Remove $\displaystyle \displaystyle 2\xi^{3/2}$ as a 'common factor' and see that the thing inside the brackets is a perfect derivative.

Alternatively, use the integrating factor technique used for solving some first order ode's.
• Nov 23rd 2010, 06:19 AM
Jester
Quote:

Originally Posted by mukmar
I'm trying to remember how to solve a type of equation I've reduced a second order partial differential to.

In the notes it goes from the equation:
$\displaystyle 2\xi u_{\xi\eta} - u_{\eta} = 0$
to
$\displaystyle 2\xi u_{\xi\eta} - u_{\eta} = 2\xi^{3/2}\frac{d}{d\xi}\left(\frac{u_{\eta}}{\xi&{1/2}}\right) = 0$

With subscripts denoting partial derivatives.

I know that these two steps are equivalent, I'm just wondering on what the procedure used is to get the second equivalent form.

If the procedure has a name which I can look up.

Any assistance would be greatly appreciated, thank you.

It might be easier to let $\displaystyle u_{\eta} = v$ so you have $\displaystyle 2 \xi v_{\xi} = v$ a separable ODE.