# Find the general solution to the equation 2u_x + u_xy = 0

• Nov 19th 2011, 12:22 PM
feyomi
Find the general solution to the equation 2u_x + u_xy = 0
where u_x is u(x,y) differentiated with respect to x and u_xy is u(x,y) twice differentiated with respect to x and with respect to y

How would I go about solving this?

Thanks for any help.
• Nov 19th 2011, 12:42 PM
feyomi
Re: Find the general solution to the equation 2u_x + u_xy = 0
I'm pretty sure I've done it now actually.

u = e^(ax+2y)
• Nov 19th 2011, 03:11 PM
Jester
Re: Find the general solution to the equation 2u_x + u_xy = 0
What you present is just one solution. Here are two more

(1) $u = e^{-2y} \sin x$

(2) $u = x e^{-2y} + y^2$

Why not let $v = u_x$ and see where that gets you.
• Nov 20th 2011, 05:40 AM
feyomi
Re: Find the general solution to the equation 2u_x + u_xy = 0
Quote:

Originally Posted by Danny

Why not let $v = u_x$ and see where that gets you.

This gives 2v + v_y = 0

I'm not really sure what to do
• Nov 20th 2011, 05:44 AM
Jester
Re: Find the general solution to the equation 2u_x + u_xy = 0
Separate and integrate noting that you'll get a function of integration.
• Nov 20th 2011, 05:45 AM
HallsofIvy
Re: Find the general solution to the equation 2u_x + u_xy = 0
Since only differentiation with respect to y is involved, you can solve that as the ODE
$\frac{dv}{dy}= -2y$. Of course, the "constant" may be a function of x. What does that give you for v? What does that make the orginal equation?

Remember that the general solution to a partial differential equation may involve unknown functions of the variables rather than unknown constants.
• Nov 20th 2011, 05:57 AM
feyomi
Re: Find the general solution to the equation 2u_x + u_xy = 0
Quote:

Originally Posted by HallsofIvy
Since only differentiation with respect to y is involved, you can solve that as the ODE
$\frac{dv}{dy}= -2y$.

surely you mean dv/dy = -2v ?
• Nov 22nd 2011, 05:15 AM
Ackbeet
Re: Find the general solution to the equation 2u_x + u_xy = 0
Quote:

Originally Posted by feyomi
surely you mean dv/dy = -2v ?

I'm sure that's what HoI meant.