# A differential equation from a practical problem

Solving a problem which relates to the movement of a charged particle in an electric field I had to solve the following diff-equation:

$$y\frac{dy}{dx}=-\frac{a}{x}+by$$ where $(a,b>0)\,\text{and}\,y(x_0)=0;x_0>0$

Wolfram Alpha is not able to solve it.

Any hint?

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$yy'=\frac12 (y^2)'$, not sure if it helps... –  draks ... Aug 31 '12 at 14:22
Why are you expecting it to be solvable? –  Sasha Aug 31 '12 at 14:26
@Sasha Is there any reason to believe otherwise? –  Martin Gales Aug 31 '12 at 14:35
at least approximately. –  Martin Gales Aug 31 '12 at 14:38
@Sasha: Doesn't the Picard–Lindelöf theorem tell us that there will always be a unique local (possibly global) solution to any first order ODE with initial conditions? –  Fly by Night Aug 31 '12 at 14:39

Here is an implicit solution derived by maple

$$\left\{ {\it \_C1}+ \left( -2\,{{\rm e}^{-1/2\,{\frac { \left( bx-y \left( x \right) \right) ^{2}}{a}}}}\sqrt {a}- {{\rm erf}\left(1/2\,{\frac { \left( bx-y \left( x \right) \right) \sqrt {2}}{\sqrt {a}}}\right)} \sqrt {2}\sqrt {\pi }bx \right) {x}^{-1}=0 \right\}\,.$$

where ${\rm erf } (x)$ is the error function

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Thank you. If y(x0)=0 then all OK? –  Martin Gales Aug 31 '12 at 15:18
If you differentiate the left hand side with respect to x and then put $x=x_0$ and $y(x_0)=0$ then you get $$\frac{2\sqrt{a}}{x_0^2}e^{-b^2x_0/2a} = 0 \, .$$ The only way this can have a solution is if $a = 0$. In fact we need $a \to 0^+.$ –  Fly by Night Aug 31 '12 at 19:56

Maybe I'm missing something here. But your initial condition $y(x_0) = 0$ implies that $a = 0$. If

$$y \frac{dy}{dx} = by - \frac{a}{x}$$

then why not substitute $x = x_0$ to give $0 = -a/x_0$? Since $x_0 > 0$ it follows that $a=0$.

$$y \frac{dy}{dx} = by$$

has a very simple solution: either $y \equiv 0$ or $y = bx + k$ for any $k \in \mathbb{R}.$ Imposing the condition that $y(x_0) = 0$ means that $k = -bx_0$ and so $y \equiv 0$ and $y(x) = b(x - x_0)$ are the two solutions. You'll need to change your initial conditions from $a,b > 0$ to $a,b \ge 0.$

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Thank you for this hint. I will try. How about special functions? Hopeless? –  Martin Gales Aug 31 '12 at 14:56
I've edited my answer while you made a comment. I think I have a solution. –  Fly by Night Aug 31 '12 at 14:58
"But your initial condition y(x0)=0 implies that a=0". I do not understand this. No, a>0. –  Martin Gales Aug 31 '12 at 15:05
Your initial conditions are inconsistent. Put $x = x_0$ into your ODE, you get $y(x_0)y'(x_0) = by(x_0) - a/x_0.$ You've told me that $y(x_0) = 0$ and so your ODE reduces to $0 = -a/x_0$. Provided $x_0 \neq 0$ $-a/x_0 = 0 \iff a = 0.$ You may have copied something down wrong. If you insist that $a>0$ then there literally is no solution because your assumptions are inconsistent and produce a contradiction. –  Fly by Night Aug 31 '12 at 15:09
At $y(x_0)=0$ i get: $y^2=2a\ln\frac{x_0}{x};x<x_0$. Near $x_0$! –  Martin Gales Aug 31 '12 at 15:30