# Thread: Partial Differential Quotation - Maple 15

1. ## Partial Differential Quotation - Maple 15

Hello!

u(x,y)

du/dx +2*x*u=0

du/dx is the partial derivate.

How do you do this in Maple?

I have tried this:
PDE:=du/dx u(x,y)+2*x*(u(x,y)=0
ans(PDE)

but get wrong answer

Kjell

2. ## Re: Partial Differential Quotation - Maple 15

u may well be a function of two variables, say, x and y, but since y does not appear in the equation, it is can be treated as an ordinary equation. I also cannot imagine why one would want to use 'Maple'. The equation du/dx= -2xu can be separated to du/u= -2xdx and then, integrating both sides, $ln(|u|)= -x^2+ C$. Now, if u is a function of x and y (or more variables) the other variables are treated as constants in differentiating with respect to x so the "constant", C may be functions of those variables. If we are given that u is a function of x and y only, then $u(x,y)= e^{-x^2+ f(y)}= e^{f(y)}e^{-x^2}$. Of course, e to the power of an unknown function of y is itself an unknown function of y so we could also have used the solution $u(x)= Ce^{-x^2}$ to get $u(x,y)= f(y)e^{-x^2}$ where f(y) is an arbitrary function of y.

What answer did Maple give?

3. ## Re: Partial Differential Quotation - Maple 15

Hello!
Thank you very much for you answer.I understand now.
Maple gave now the same: f(x,y)=F(y)*e^-x^2

Kjell