A simple Partial Differential Equation? I thought... I was looking at the Partial Differential Equation involving function:
$$ z(x,y)$$ 
$$ \frac{\partial z}{\partial x} + c \frac{\partial z}{\partial y} = 0 $$
Which fairly intuitively has a solution:
$$ z = f\left( x - \frac{1}{c} y \right)$$
Now I was considering the generalization of this equation to:
$$ \frac{\partial z}{\partial x} + (c_1x + c_2)  \frac{\partial z}{\partial y} = 0 $$
But instead of starting too large I would focus first on:
$$ \frac{\partial z}{\partial x} + c_1 x\frac{\partial z}{\partial y} = 0 $$
Which itself i focused on:
$$ \frac{\partial z}{\partial x} + x\frac{\partial z}{\partial y} = 0 $$
I did some algebraic debauchery and have concluded that:
$$ g = a_0 + a_1 x - 2 a_1 x\sum_{i=1}^{\infty} \left[\frac{1}{i}\begin{pmatrix} 2i-2 \\ i-1 \end{pmatrix} \left( \frac{y}{2x^2} \right)^n \right] $$ 
Is a solution to this. But I do not know its closed form.
Work So Far:
One notes that the solution $ z = f\left( x - \frac{1}{c} y \right)$ to the first problem can be derived by noting that:
$$ z = 1, x - \frac{1}{c}y, \left(x - \frac{1}{c}y\right)^2 ... $$
Are all solutions and the transform is linear so any function of the form:
$$ F = a_0(1) + a_1 \left(x - \frac{1}{c}y \right) + a_2 \left(x - \frac{1}{c}y\right)^2 ... $$
Is a solution, which is the general power series of all functions of the form
$$ f \left( x - \frac{1}{c}y\right) $$
Using that same intuition note:
$$ g_0 = a_0 $$ 
Is a solution... then we can consider $$ g_1 = a_0 + a_1 x $$ and ask what term should be added to $g_1$ to make $g_2$ such that $$ \frac{\partial g_1}{\partial x} + x \frac{\partial g_2}{\partial y} = 0 $$. Then recursively generate $g_3, g_4 ... $
By doing so we conclude that one solution is 
$$ g = a_0 + a_1 x - 2 a_1 x\sum_{i=1}^{\infty} \left[\frac{1}{i}\begin{pmatrix} 2i-2 \\ i-1 \end{pmatrix} \left( \frac{y}{2x^2} \right)^n \right] $$ 
What is the closed form for this guy? From here it becomes clear that if we consider $g^r$
$$ \frac{\partial g^r}{\partial x} + x \frac{\partial g^r}{\partial y} =  $$ 
$$ r g^{r-1} \left( \frac{\partial g}{\partial x} + x \frac{\partial g}{\partial y} \right)  = r g^{r-1}(0) = 0 $$
And since our transform is linear:
$$ z = e_0 + e_1g + e_2 g^2  + e_3 g^3 ...  $$ 
Is a solution to the equation meaning that the general solution is 
$$ z = f(g) $$ 
 A: For $\dfrac{\partial z}{\partial x}+(c_1x+c_2)\dfrac{\partial z}{\partial y}=0$ ,
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{dy}{dt}=c_1x+c_2=c_1t+c_2$ , letting $y(0)=y_0$ , we have $y=\dfrac{c_1t^2}{2}+c_2t+y_0=\dfrac{c_1x^2}{2}+c_2x+y_0$
$\dfrac{dz}{dt}=0$ , letting $z(0)=f(y_0)$ , we have $z(x,y)=f(y_0)=f\left(y-\dfrac{c_1x^2}{2}-c_2x\right)$
A: After playing around with wolfram alpha I found that
$$ \sum_{i=1}^{\infty} \left[ \frac{1}{i} \begin{pmatrix} 2i-2 \\ i-1 \end{pmatrix}  u^i \right] = \frac{1}{2} (1 - \sqrt{1 - 4u} )$$
Therefore
$$g = a_0 + a_1 x - 2a_1 x \frac{1}{2} \left(1 - \sqrt{1 - 4\frac{y}{2x^2}}  \right)  = $$
$$ a_0 +  a_1x - a_1x \left(1 - \sqrt{1 - 2 \frac{y}{x^2}} \right) = $$
$$ a_0 + a_1x - a_1x \left(1 - \frac{\sqrt{x^2 - 2 y}}{x} \right)  = $$
$$ a_0 + a_1x - a_1 \left(x - \sqrt{x^2 - 2 y} \right)  = $$
$$ a_0 + a_1 \sqrt{x^2 - 2y}$$
We verify:
$$ \frac{\partial g}{\partial x} = a_1 \frac{x}{\sqrt{x^2 - 2y}}$$
$$ x\frac{\partial g}{\partial y} = -a_1 \frac{x}{\sqrt{x^2 - 2y}}$$
$$\frac{\partial g}{\partial x} + x\frac{\partial g}{\partial y} = 0 $$
