Thread: 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

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, . 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 . 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 to get where f(y) is an arbitrary function of y.

What answer did Maple give?

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