How to continue to find a solution to the PDE? I want to find a solution $u(x,y)$ for $\begin{cases}1 \cdot \frac{\partial u}{\partial x}+ (-x) \cdot \frac{\partial u}{\partial y}- x \cdot u = 0 \\u(x,-x^2)=1
\end{cases}$
I wanted to try it like in 
http://www1.maths.leeds.ac.uk/~kersale/Teach/M3414/Notes/m3414_1.pdf
on page $11$ the example.
We put $a=1 ,\quad b=-x ,\quad c=-x ,\quad f=0$ and get the characteristic euqation $$\frac{dy}{dx}=\frac{-x}{1}=-x$$
Solving the ODE gives us $$y=-\frac{1}{2}x^2+c \quad c \in \Bbb R$$
Let $\eta$ be a constant,so $$\eta=\eta(x,y)=y+\frac{1}{2}x^2$$
now choose $x=\xi$ and get $$\eta=y+\frac{1}{2}\xi^2 \Rightarrow y= \eta-\frac{1}{2}\xi^2$$
Put $w(\eta,\xi)=u(x,y)$ and get $$\frac{\partial u}{\partial x}=\frac{\partial w}{\partial \xi}+\frac{\partial w}{\partial \eta}\cdot x=\frac{\partial w}{\partial \xi}+\xi \cdot\frac{\partial w}{\partial \eta}$$
$$\frac{\partial u}{\partial y}=0+\frac{\partial w}{\partial \eta}\cdot 1=\frac{\partial w}{\partial \eta}$$
our PDE becomes $$(\frac{\partial w}{\partial \xi}+\xi \cdot\frac{\partial w}{\partial \eta})-\xi \frac{\partial w}{\partial \eta}-\xi \cdot w=0$$
which we can write as $$\frac{\partial w}{\partial \xi} -\xi \cdot w=0$$
Can I say that $w=e^{1/2\xi^2}$ ? or how do I get the soultion to the original PDE ? 
 A: Two things 
First dont forget the function of $\eta$
$$\frac{\partial w}{\partial \xi} -\xi \cdot w=0$$
$$\int \frac{d w}{w} =\int \xi d\xi$$
$$\ln|w|=\frac {\xi ^2}{2}+h(\eta)$$
$$....$$
2
You have $\eta$ in function of x,y
$$\implies \eta= y+\frac {x^2}2$$
And also $\xi$ in function of x y
$$\implies \xi =x$$
Use them to get back to your original variables..
A: $$\frac{\partial u}{\partial x}+ (-x) \frac{\partial u}{\partial y}= x u $$
$$\frac{dx}{1}=\frac{dy}{-x}=\frac{du}{xu}$$
First characteristics, from $\quad\frac{dx}{1}=\frac{dy}{-x}$ :
$$ 2y+x^2=c_1$$
Second characteristics, from $\quad\frac{dy}{-x}=\frac{du}{xu}$ :
$$u\,e^y=c_2$$
General solution of the PDE :
$$u\,e^y=F(2y+x^2)$$
$F$ is an arbitrary function, to be determined according to the boundary condition.
$$u=e^{-y}F(2y+x^2)$$
Condition : $\quad u(x,-x^2)=1=e^{-(-x^2)}F(2(-x^2)+x^2)$
$$F(-x^2)=e^{-x^2}$$
Let $\quad X=-x^2$
$$F(X)=e^{X}$$
Now, the function $F$ is determined. We put it into the above general solution where $X=2y+x^2$ , then $F(2y+x^2)=e^{2y+x^2}$ .
$u(x,y)=e^{-y}F(2y+x^2)=e^{-y}e^{2y+x^2}$
$$u(x,y)=e^{y+x^2}$$
$\\$
IN ADDITION, comment about your calculus :
$$\frac{\partial w}{\partial \xi} -\xi \cdot w=0$$
$$w(\eta,\xi)=f(\eta)e^{\xi^2/2}$$
$$u(x,y)=f(y+\frac{x^2}{2})e^{x^2/2}$$
Condition :$\quad u(x,-x^2)=1=f(-x^2+\frac{x^2}{2})e^{x^2/2}=f(-\frac{x^2}{2})e^{x^2/2}$
Let $X=-x^2/2\quad;\quad 1=f(X)e^{-X} \quad;\quad f(X)=e^{X} \quad$ that we put into the general solution where  $X=y+\frac{x^2}{2}$ , then $f(y+\frac{x^2}{2})=e^{y+\frac{x^2}{2}}$
$$u(x,y)=e^{y+\frac{x^2}{2}}e^{x^2/2}=e^{y+x^2}$$
