How to solve a Partial Differential Equation I have not solved PDEs in eons.
I am battling to solve the following:
$ x^2 \frac{\partial s}{\partial x} +xy\frac{\partial s}{\partial y} =1$
I have managed to solve a similar PDE which equals zero, but I don't know how to deal with the 1 in this equation.
If the steps are similar then I would divide by $x^2$ to get $\frac{\partial s}{\partial x} +\frac{y}{x}\frac{\partial s}{\partial y} =\frac{1}{x^2}$
However I don't know where to go from here :(
NOTE: This equation is actually part of a pair of equations that I need to solve in order to determine the canonical coordinates $r(x,y)$ and $s(x,y)$.  I'm trying to solve y''=0 using symmetry.  I've solved $r(x,y)=\frac{y}{x}$ and now need to get $s(x,y)$.  
 A: I will follow your approach of dividing through by $x^{2}$. Using the method of characteristics, we obtain the ODEs
\begin{align}
\frac{dy}{dx} &= \frac{y}{x} \quad (1) \\
\frac{ds}{dx} &= \frac{1}{x^{2}} \quad (2) \\
\end{align}
Solving $(1)$ yields
\begin{align}
y &= y_{0} x \implies y_{0} = \frac{y}{x} \\
\end{align}
Solving $(2)$ yields
\begin{align}
s &= \frac{-1}{x} + f(y_{0}) \\
&= \frac{-1}{x} + f \left( \frac{y}{x} \right)
\end{align}
where $f$ is an arbitrary differentiable function. You can check via differentiation that this is the general solution. 
I would also suggest trying to use Hans' idea and change your problem to polar coordinates.
A: $$ x^2 \frac{\partial s}{\partial x} +xy\frac{\partial s}{\partial y} =1$$
System of characteristic ODEs : $\quad\frac{dx}{x^2}=\frac{dy}{xy}=\frac{ds}{1}$
First set of characteristics, from $\quad\frac{dx}{x^2}=\frac{dy}{xy} \quad\implies\quad \frac{y}{x}=c_1$
Second set of characteristics, from $\quad\frac{dx}{x^2}=\frac{ds}{1} \quad\implies\quad s+\frac{1}{x}=c_2$
General solution on implicit form : $\quad F\left(\frac{y}{x} \:,\:s+\frac{1}{x} \right)=0\quad$ where $F$ is any differentiable function of two variables.
Or equivalently, solution on explicit form : $\quad s+\frac{1}{x}=f\left(\frac{y}{x} \right)\quad$ where $f$ is any differentiable function.
$$s=-\frac{1}{x}+f\left(\frac{y}{x} \right)$$
A: We start by rewriting the equation, 
$$ x \frac{\partial s}{\partial x } + y\frac{\partial s }{\partial y} = \frac{1}{x}$$
in polar coordinates this becomes, 
$$ r \cos\theta \frac{\partial s}{\partial x } + r\sin\theta \frac{\partial s }{\partial y} = \frac{1}{r\cos\theta}$$
Note that, 
$$ \frac{\partial}{\partial r } = \cos\theta \frac{\partial}{\partial x} + \sin\theta \frac{\partial}{\partial y} ,$$
which allows us to rewrite our equation as, 
$$ r\frac{\partial s}{\partial r} = \frac{1}{r\cos\theta}$$
$$ \frac{\partial s}{\partial r} = \frac{1}{r^2\cos\theta}$$
integrating with respect to $r$ we get, 
$$ s  = -\frac{1}{r\cos\theta} + f(\theta) $$
$$ s  = -\frac{1}{x} + f(\theta) $$
Where $f$ is an arbitrary function of $\theta$. We can write replace this with an arbitrary function of $y/x$; this is because a function of $\theta$ can also be written as a function of $\tan\theta = y/x$. We will call this new function $F$, where $F(y/x)=f(\theta)$. 
$$ s  = -\frac{1}{x} + F(y/x) $$
