Partial Diff qn

by fredrick08
Tags: diff, partial
 P: 376 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: $$\partial$$2u/$$\partial$$x$$\partial$$y=u [b]3. The attempt at a solution[/] =>$$\int$$$$\int$$u dx dy =uxy=f(x)f(y) ???
 P: 376 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..
 HW Helper Sci Advisor Thanks P: 24,460 Set u(x,y)=f(x)*g(y). What is your partial derivative in terms of f(x), f'(x), g(y), and g'(y)?
P: 376

Partial Diff qn

so $$\partial$$2u/$$\partial$$x$$\partial$$y=f(x)g(y)??
so in terms of f'(x) and g'(y) im not sure..
 HW Helper P: 1,495 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 $$\frac{\partial^2 f(x)g(y)}{\partial x \partial y}=f(x)g(y)$$. Now what Dick asks you to do is to calculate $$\frac{\partial^2 f(x)g(y)}{\partial x \partial y}$$. Hint: $$\frac{d}{dx}f(x)= ?$$ In terms of f'(x)
 HW Helper Sci Advisor Thanks P: 24,460 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.
 HW Helper P: 1,495 Sorry, that was sloppy of me I actually meant f(x)g(x). I will fix it.
 P: 376 hmm d/dx of f(x).. isn't that just f'(x)??.... is it something like the integral of f'(x)g'(y)??
 P: 376 so... $$\frac{\partial^2 f(x)g(y)}{\partial x \partial y}$$ =$$\int$$$$\int$$f'(x)g'(y) dx dy??
HW Helper
P: 1,495
What you've written down now is the equivalent to $\frac{d}{dx}f(x)=\int f'(x) dx$. Which is wrong. Lets test it, f(x)=x then $\frac{d}{dx}x=1 \neq \int 1 dx=x+c$.

 hmm d/dx of f(x).. isn't that just f'(x)??
This is right, so if you evaluate $$\frac{\partial^2 f(x)g(y)}{\partial x \partial y}$$ you get....?
 P: 376 u get f'(x)g'(y)?? Im confused becasue when u take partial of f(x), u get d/dy(d/dx*f(x))=0??
 HW Helper P: 1,495 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). $$\frac{\partial^2 f(x)g(y)} {\partial x \partial y}=\frac{\partial}{\partial x} f(x) \frac{\partial}{\partial y}g(y)=f'(x)g'(y)$$
 P: 376 ok so since f(x)g(y)=f'(x)g(y)=u(x,y)=??
 HW Helper P: 1,495 You forgot a prime it should be $f'(x)g'(y)=f(x)g(y)=u(x,y)$. You can instantly see which function satisfies the equation, but you can also separate the variables.
 P: 376 e^x+y?? so f(x)=e^x and g(y)=e^y same as the derivativeS?
 HW Helper P: 1,495 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?
 P: 376 e^xy?? so f(x)=e^x and g(y)=e^y same as the derrivativeS? srry this was an accident
 HW Helper P: 1,495 Don't randomly guess, $f'(x)g'(y)=y e^{xy}xe^{xy} \neq e^{xy}$. Separate the variables first, that is all the x dependent stuff on one side and all the y dependent stuff on the other side.

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