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

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- Jul 20th 2012, 05:57 AMkjellPartial 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 - Jul 20th 2012, 07:25 AMHallsofIvyRe: 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, $\displaystyle ln(|u|)= -x^2+ C$. 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 $\displaystyle u(x,y)= e^{-x^2+ f(y)}= e^{f(y)}e^{-x^2}$. 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 $\displaystyle u(x)= Ce^{-x^2}$ to get $\displaystyle u(x,y)= f(y)e^{-x^2}$ where f(y) is an arbitrary function of y.

What answer did Maple give? - Jul 20th 2012, 08:32 AMkjellRe: 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