1. The problem statement, all variables and given/known data Does the following differential equation for u(x,y) have solutions which have the form of a product of functions of each independent variable: [tex]\partial[/tex]^{2}u/[tex]\partial[/tex]x[tex]\partial[/tex]y=u 3. The attempt at a solution[/] =>[tex]\int[/tex][tex]\int[/tex]u dx dy =uxy=f(x)f(y) ???
im not sure if that even makes sense, but if u(x,y)=f(x)f(y) and d*f(x)/dx=u/constant and d*f(y)/dy=constant... i think, i don't know im confused..
so [tex]\partial[/tex]^{2}u/[tex]\partial[/tex]x[tex]\partial[/tex]y=f(x)g(y)?? so in terms of f'(x) and g'(y) im not sure..
You have an equation that equates the derivative of u to u itself. You can already guess what u will be, but lets not guess. Setting u(x,y)=f(x)g(y), then the equation becomes [tex]\frac{\partial^2 f(x)g(y)}{\partial x \partial y}=f(x)g(y)[/tex]. Now what Dick asks you to do is to calculate [tex]\frac{\partial^2 f(x)g(y)}{\partial x \partial y}[/tex]. Hint: [tex] \frac{d}{dx}f(x)= ? [/tex] In terms of f'(x)
I actually meant u(x,y)=f(x)g(y). There's no reason why the function of x would have to be the same as the function of y. And yes, evaluate the derivative.
so... [tex] \frac{\partial^2 f(x)g(y)}{\partial x \partial y} [/tex] =[tex]\int[/tex][tex]\int[/tex]f'(x)g'(y) dx dy??
What you've written down now is the equivalent to [itex]\frac{d}{dx}f(x)=\int f'(x) dx[/itex]. Which is wrong. Lets test it, f(x)=x then [itex]\frac{d}{dx}x=1 \neq \int 1 dx=x+c[/itex]. This is right, so if you evaluate [tex]\frac{\partial^2 f(x)g(y)}{\partial x \partial y}[/tex] you get....?
Yes that's correct, but you're not taking the partial of f(x), you're talking the partial derivative of f(x)g(y). [tex] \frac{\partial^2 f(x)g(y)} {\partial x \partial y}=\frac{\partial}{\partial x} f(x) \frac{\partial}{\partial y}g(y)=f'(x)g'(y) [/tex]
You forgot a prime it should be [itex]f'(x)g'(y)=f(x)g(y)=u(x,y)[/itex]. You can instantly see which function satisfies the equation, but you can also separate the variables.
That would be a solution indeed, you're on the right track. it is not the general solution however. The general solution will involve integration constants and a constant in the argument of the exponential. Can you see how to integrate the differential equation?
Don't randomly guess, [itex]f'(x)g'(y)=y e^{xy}xe^{xy} \neq e^{xy}[/itex]. Separate the variables first, that is all the x dependent stuff on one side and all the y dependent stuff on the other side.
If c are constants then f'(x)=g'(y)=0, which doesn't fit either. Can you please separate the variables first.