Partial differential equation

**mukmar** · November 22nd 2010, 08:49 PM

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:
$2\xi u_{\xi\eta} - u_{\eta} = 0$

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

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.

**BobP** · November 23rd 2010, 02:06 AM

Remove $\displaystyle 2\xi^{3/2}$ 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.

**Danny** · November 23rd 2010, 06:19 AM

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

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 $u_{\eta} = v$ so you have $2 \xi v_{\xi} = v$ a separable ODE.

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