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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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    Similar to http://math.stackexchange.com/questions/14638012016-05-22

2 Answers 2

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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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    @MartinGales: It seems you've got mixed up with variables and functions. Some times you see $x$ representing position, but $x$ is$a$function of $t$. In your problem $x$ plays the roles of time while $y(x)$ is position. Look at Hooke's law: $m\ddot{x} = -kx$ Here $t$ is time, $x(t)$ is the position at time $t$, $\dot{x}(t)$ is the velocity at time $t$ and $\ddot{x}(t)$ is acceleration at time $t$. In your notation, this would be $my'' = -ky$ Where $x$ is time, $y(x)$ is the position at time $x$, $y'(x)$ is the velocity at time $x$ and $y''(x)$ is acceleration at time $x$.2012-09-02
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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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    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^+.$2012-08-31