# Partial differentail equation

1. Dec 20, 2009

### Dell

X*U'x+Y*U'y=U
{U=1-x2}
{y=1}

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dx/x=dy/y=du/u

$$\int$$dx/x=$$\int$$dy/y

ln|x|=ln|y|+ln|c1|
x/y=c1

$$\int$$dx/x=$$\int$$du/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

2. Dec 20, 2009

### rock.freak667

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

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 ?

3. Dec 20, 2009

### HallsofIvy

Staff Emeritus
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.