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.

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- Nov 19th 2011, 11:22 AMfeyomiFind 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, 11:42 AMfeyomiRe: 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, 02:11 PMJesterRe: Find the general solution to the equation 2u_x + u_xy = 0
What you present is just one solution. Here are two more

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

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

Why not let $\displaystyle v = u_x$ and see where that gets you. - Nov 20th 2011, 04:40 AMfeyomiRe: Find the general solution to the equation 2u_x + u_xy = 0
- Nov 20th 2011, 04:44 AMJesterRe: 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, 04:45 AMHallsofIvyRe: 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

$\displaystyle \frac{dv}{dy}= -2y$. 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, 04:57 AMfeyomiRe: Find the general solution to the equation 2u_x + u_xy = 0
- Nov 22nd 2011, 04:15 AMAckbeetRe: Find the general solution to the equation 2u_x + u_xy = 0