# Partial Diff qn

## Homework Statement

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

3. The Attempt at a Solution [/]

=>$$\int$$$$\int$$u dx dy =uxy=f(x)f(y) ???

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## Answers and Replies

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..

Dick
Science Advisor
Homework Helper
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)?

so $$\partial$$2u/$$\partial$$x$$\partial$$y=f(x)g(y)??
so in terms of f'(x) and g'(y) im not sure..

Cyosis
Homework Helper
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)

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Dick
Science Advisor
Homework Helper
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.

Cyosis
Homework Helper
Sorry, that was sloppy of me I actually meant f(x)g(x). I will fix it.

hmm d/dx of f(x).. isn't that just f'(x)??.... is it something like the integral of f'(x)g'(y)??

so... $$\frac{\partial^2 f(x)g(y)}{\partial x \partial y}$$ =$$\int$$$$\int$$f'(x)g'(y) dx dy??

Cyosis
Homework Helper
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....?

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??

Cyosis
Homework Helper
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).

$$\frac{\partial^2 f(x)g(y)} {\partial x \partial y}=\frac{\partial}{\partial x} f(x) \frac{\partial}{\partial y}g(y)=f'(x)g'(y)$$

ok so since f(x)g(y)=f'(x)g(y)=u(x,y)=??

Cyosis
Homework Helper
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.

e^x+y?? so f(x)=e^x and g(y)=e^y same as the derivativeS?

Cyosis
Homework Helper
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?

e^xy?? so f(x)=e^x and g(y)=e^y same as the derrivativeS?

srry this was an accident

Cyosis
Homework Helper
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.

na im confused now... like f(x)=e^c1 g(y)=e^c2 and u(x,y)=e^c1+c2??

Cyosis
Homework Helper
If c are constants then f'(x)=g'(y)=0, which doesn't fit either. Can you please separate the variables first.

im confused because when u take the derivative of a constant its zero?

Cyosis
Homework Helper
Yes when you take the derivative of a constant it's zero. You may want to elaborate on your confusion.

oh ok i will try we only just learned this... U(x,y)=F(x)G(y)

F'(x)G'(y)=F(x)G(y) divide by U(x,y)=F'(x)G'(y)/F(x)G(y)=1
im not sure the ones i have done arent as confusing as this one

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no thats not rite... umm F'/F=G/G'=gamma?

Cyosis
Homework Helper
No you have the y and x terms on the same side now. Separation of variables means that you put all the functions depending on x on one side and all functions depending on y on the other side. Can you do this?