Consider the equation $u_x + 3x^2y^2u_y = 0$ :
Find a solution satisfying $u(x, 0) = 1/(1+x)$, or explain why no such solution exists.
My attempt:
I've managed to solve the equation and find the general solution as $u(x,y)=f(x^3 + 1/y)$.
Now I know $u(x,0) = f(x^3)$, but I'm not sure what method to use further to show that solutions exist or don't exist.
When we know that $u(x,y)$ is in terms of $f(z)$ , where $z=x^3+1/y$ , we already know that $z$ is not defined at $y=0$ (due to $1/0$) , hence $f$ is not defined there & eventually $u$ is not defined there.
In other words , there is no Solution where $y=0$ occurs.
[[ Check that $u(x,0)=f(x^3+1/0) \not = f(x^3)$ ]]
In Case the Criteria given had been $u(x,1)=1/(1+x)$ , we can get Solutions like $u(x,y)=(1+\sqrt[3]{(x^3+1/y-1)})^{-1}$.
It will become $u(x,1)=(1+\sqrt[3]{(x^3+1/1-1)})^{-1}=1/(1+x)$.
In Case the PDE Solution had been $u(x,y)=f(x^3+y)$ , we can make it $u(x,0)=f(x^3+0)=f(x^3)$ to get Solutions.
Unfortunately , what we have here involves $1/0$ , hence there are no Solutions.