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Linear partial equations

Posted September 3rd, 2007 by vic777
in

Question Details:
How can I find a solution for this eq. ?

$ (1+x^2)u_x+u_y=0 $

‹ Linear partial differential eqn. How to prove this inequality? ›

Comments

Suppose you have a solution

On September 3rd, 2007 Structure says:

Suppose you have a solution u . It has a graph G={x,y,u(x,y)}
You cut it with a plane z=c.
A section in the graph (which is a surface) is a curve, u(x,y)=c.
If you can apply the implicit function theorem you have
$ y'(x)=-\frac{u_x}{u_y} $ and from the given equation
$ y'(x)=-\frac{u_x}{u_y}=\frac{1}{1+x^2} $
So $ y(x)=\arctan(x)+C $
then $ u(x,y)=y-\arctan(x)=C $
Now you can verify that any two solutions are functional dependent which means that any solution $ u(x,y)=F(y-\arctan(x)) $

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