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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

$$ z = f(g) $$

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  • $\begingroup$ Not that it directly answers your question about the closed form of that guy, but if you change variables to $u=x$, $v=y/2x^2$, you'll find that $z=f(v)=f(y/2x^2)$ is the general solution to $z_x+xz_y=0$ (in the region $x>0$, say). $\endgroup$ – Hans Lundmark Jan 5 '15 at 12:37
  • $\begingroup$ (By the way, in your approach, I think you need $a_1=0$, otherwise $g_1$ isn't a solution...) $\endgroup$ – Hans Lundmark Jan 5 '15 at 12:38
  • $\begingroup$ $g_1, g_2 .... $ all aren't solutions, $g_{i+1}$ is simply a correction to $g_{i}$ but in the limit to infinity they form the infinite sum $\endgroup$ – frogeyedpeas Jan 5 '15 at 14:04
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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 $z(x,y)=f(y_0)=f\left(y-\dfrac{c_1x^2}{2}-c_2x\right)$

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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:

$$ \frac{\partial g}{\partial x} = a_1 \frac{x}{\sqrt{x^2 - 2y}}$$ $$ x\frac{\partial g}{\partial y} = -a_1 \frac{x}{\sqrt{x^2 - 2y}}$$

$$\frac{\partial g}{\partial x} + x\frac{\partial g}{\partial y} = 0 $$

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  • $\begingroup$ This is only defined however if $\left|\frac{y}{2x^2} \right|< \frac{1}{4}$ $\endgroup$ – frogeyedpeas Jan 5 '15 at 14:17
  • $\begingroup$ Except, even if that property is violated, if one allows complex roots then the equation is still well defined $\endgroup$ – frogeyedpeas Jan 5 '15 at 14:20

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