Finding solution of the PDE : An Initial value problem I've been given a PDE of the form $$xu_x+(x^2+y)u_y=1-\left(\frac{y}{x}-x\right)u~~; ~~u(1,y)=0$$
Attempt : Firstly it's a first order linear PDE which has the general form $$a(x,y)u_x+b(x,y)u_y=c(x,y)u+d(x,y)$$Where $a,b,c,d \in \mathcal{C}^1(\Omega)$, where $\Omega \subseteq \mathbb{R}^2$ (open and connected) and we also have $a^2+b^2 \neq 0$ on $\Omega_2$. Then first of all the largest $\Omega_2$ on which $F:=a^2+b^2=x^2+(x^2+y)\neq 0$ is : $(x,y) \in \Omega_2\setminus \{(0,0)\}$ right?
We have $s \mapsto \Gamma_s:=(f(s_0):=1,g(s_0):=s,h(s_0):=0))$. Now, from transversality conditions we get :$$J:=\mathrm{det}\, \begin{pmatrix}a&f'\\b&g'\end{pmatrix}\Bigg{|}_{f(s_0),g(s_0),h(s_0)}=\mathrm{det}\begin{pmatrix}1&0\\1+s&1\end{pmatrix}=1$$
So we atleast expect solutions to exist $\textit{locally}$ for all $s \in (-\delta,\delta),\delta>0$.
Now to find the solutions explicitly i'll employ the method of characteristics: \begin{align}\frac{\mathrm{d}x}{\mathrm{d}t}&=x~~;~x(0,s):=1 \implies x(t,s):=e^t\\\frac{\mathrm{d}y}{\mathrm{d}t}&=x^2+y~~;~ y(0,s):=s \\&=y+e^{2t} \end{align}
Solving this we get
\begin{align}   y(t,s)&=e^{2t}+c_2e^{t}~~;~y(0,s):=s \implies 
y(t,s)=e^{t}\left(e^t+s-1\right)\end{align} Lastly \begin{align}\frac{\mathrm{d}z}{\mathrm{d}t}&=1-\left(\frac{y}{x}-x\right)z~~;~z(0,s):=0 \\&=1-\left(\frac{e^t(e^t+s-1)}{e^t}-e^t\right)z=1+(1-s)z\end{align}
Which gives me $$z(t,s)=\frac{1}{s-1}\left(1-e^{t(1-s)}\right)$$
As the Jacobian was non-zero by inverse function theorem i can expect that i can invert the map $$(t,s)\mapsto \left(x(t,s),y(t,s)\right)$$ and i can solve $x=x(t,s),y=y(t,s)$ uniquely for $t$ and $s$ and the map $$(x,y)\mapsto \left(T(x,y);S(x,y)\right)$$ should be of class $\mathcal{C}^1$.
So i have $x(t,s)=e^t \implies t=\ln\,x=:T(x,y)$ and $y(t,s)=e^{\ln\,x}\left(e^{\ln \,x}+s-1\right)=x(x+s-1)$ i.e $s=(y/x)-x+1=:S(x,y)$. From here i can write $$u=z(T(x,y),S(x,y))=\frac{x}{y-x^2}\left(1-x^{\frac{x^2-y}{x}}\right)$$
Which is valid when $y>x^2$. I'm not sure about my workings here, appreciate any hints and help. Thanks.
 A: I agree with your result as shown below.
$$xu_x+(x^2+y)u_y=1-(\frac{y}{x}-x)u$$
Charpit-Lagrange system of characteristic ODEs :
$$\frac{dx}{x}=\frac{dy}{x^2+y}=\frac{du}{1-(\frac{y}{x}-x)u}$$
A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{x^2+y}$ which is a first order linear ODE :
$$\frac{y}{x}-x=c_1$$
A second characteristic equation comes from solving $\frac{dx}{x}=\frac{du}{1-(\frac{y}{x}-x)u}=\frac{du}{1-c_1u}$ which is a first order linear ODE :
$$x^{c_1}(u-\frac{1}{c_1})=c_2$$
The general solution of the PDE (any boundary condition) expressed on the form of implicit equation $c_2=F(c_1)$ with arbitrary function $F$ is :
$$x^{c_1}(u-\frac{1}{c_1})=F(c_1)$$
$$x^{\frac{y}{x}-x}(u-\frac{1}{\frac{y}{x}-x})=F\left(\frac{y}{x}-x\right)$$
$$\boxed{u(x,y)=\frac{1}{\frac{y}{x}-x}+x^{-\frac{y}{x}+x}F\left(\frac{y}{x}-x\right)}$$
Determination of the function $F$ in order to satisfy the condition $u(1,y)=0$ :
$$u(1,y)=0=\frac{1}{y-1}+F\left(y-1\right)$$
Let $X=y-1 \quad;\quad y=X+1$
$$0=\frac{1}{X}+F\left(X\right)$$
$$F(X)=-\frac{1}{X}$$
Now the function $F(X)$ is known. we put it into the above general solution where $X=\frac{y}{x}-x$
$F\left(\frac{y}{x}-x\right)=-\frac{1}{\frac{y}{x}-x}$
The particular solution which satisfies both the PDE and the condition is :
$${u(x,y)=\frac{1}{\frac{y}{x}-x}-x^{-\frac{y}{x}+x}}\frac{1}{\frac{y}{x}-x}$$
$$\boxed{{u(x,y)=\frac{x}{y-x^2}}\left(1 -x^{\frac{x^2-y}{x}}\right)}$$
