Partial Differential Equations Help

[walter9459]

Homework Statement

Show that u=f(2x+y^2)+g(2x-y^2) satisfies the equation y^2 d^2u/dx^2 + (1/y) du/dy - d^2u/dy^2=0 where f and g are arbitrary (twice differentiable) functions.

Homework Equations

The Attempt at a Solution

I came up with fxx=0 fyy=2 gxx=0 gyy= 2. But didn't know if this was correct and if so what to do next. I do not want the solution but how to go about solving this problem. Thanks!

 

[Dick]

You want to find things like uxx, uy and uyy to substitute into your equation. What are those? Oh, and fxx is not 0. E.g. fx=f'(2x+y^2)*2. Use the chain rule.

 

[walter9459]

I am still confused, I do not see where the chain rule comes into play, f(x,y) = 2x + y². So why isn't fx equal to 2? I do not see how you are coming up with (2x + y²)*2. I desperately need to understand this whole concept! Thanks!

 

[Dick]

Chain rule. f(x,y) isn't equal to 2x+y^2. It's equal to f(2x+y^2). The x derivative is f'(2x+y^2)*2.

 

[walter9459]

Sorry to be a pain but I think I am finally getting the idea and want to make sure this is true. ux = f'(2x+y²)*2 + g'(2x-y²)*2
uy = f'(2x+y²)*2y + g'(2x-y²)*-2y
uxx = f''(2x+y²)*0 + g''(2x-y²)*0
uyy = f''(2x+y²)*2 + g''(2x-y²)*-2

then I just plug these results into the equation y²uxx + (1/y) uy - uyy = 0. Is that correct?

Thanks!

 

[Dick]

ux and uy are ok. Not uxx and uyy. fxx=(f'(2x+y^2)*2)'=2*(f'(2x+y^2))'=4*f''(2x+y^2). The '2' is a multiplicative constant. You need to use the product rule on 2y*f'(2x+y^2), since it's the product of two functions of y. Yes, then just plug them into your equation.