What is the solution to the partial differential equation with given conditions?

[Dell]

X*U'_x+Y*U'_y=U
{U=1-x²}
{y=1}
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dx/x=dy/y=du/u

\intdx/x=\intdy/y

ln|x|=ln|y|+ln|c₁|
x/y=c₁

\intdx/x=\intdu/u

ln|x|=ln|u|+ln|c₂|
x/u=c₂

\Phi(x/u , x/y)=0

x/u=\phi(x/y)

now i use the conditions
{U=1-x²}
{y=1}

x/(1-x²)=\phi(x)=x/u

and all i get is U=1-x² but i already know that

the correct answer is u=(y² -x²)/y

 

[rock.freak667]

X*U'_x+Y*U'_y=U
{U=1-x²}
{y=1}

Is your equation

x \frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=1-x^2

or


x \frac{\partiall u}{\partial x}+y\frac{\partial u}{\partial y}=0

with u(x,0)=1-x² and u(0,y)=1 ?

 

[HallsofIvy]

I interpret this to mean that U(x,y) satisfies
]x\frac{\partial U}{\partial x}+ \frac{\partial U}{\partial y}= U
with the condition that U(x, 1)= 1- x^2

But then it is peculiar that there is no condition for x a constant.