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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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    $yy'=\frac12 (y^2)'$, not sure if it helps...2012-08-31
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    Why are you expecting it to be solvable?2012-08-31
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    @Sasha Is there any reason to believe otherwise?2012-08-31
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    at least approximately.2012-08-31
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    @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?2012-08-31
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    @FlybyNight Sorry, I misspoke. I meant to ask "why are you expecting it to be solvable in closed-form".2012-08-31
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    A solution in terms of special functions is also a closed form for me2012-08-31
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    I think I've found a solution. See below.2012-08-31
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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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    Thank you for this hint. I will try. How about special functions? Hopeless?2012-08-31
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    I've edited my answer while you made a comment. I think I have a solution.2012-08-31
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    "But your initial condition y(x0)=0 implies that a=0". I do not understand this. No, a>0.2012-08-31
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    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.2012-08-31
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    At $y(x_0)=0$ i get: $y^2=2a\ln\frac{x_0}{x};x$x_0$! – 2012-08-31
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    What can happen is $y'(x_0) = +\infty$.2012-08-31
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    @MartinGales: There is no solution with the initial conditions that you have given. You need to sit down and think about this in more depth. Let's look at your solution. If $x$x_0$! If $y^2 = 2a\ln(x_0/x)$ differentiating gives $2yy' = -2a/x$. If you let $x \to x_0$ then $y \to 0$ and so $a \to 0$. Your own solution is hinting that $a = 0.$ Please, spend some time thinking about this. I can assure you that if $y(x_0) = 0$ the $a$ must be zero! – 2012-08-31
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    @MartinGales: Remember my series solution, in which I did not assume $a=0$? You have $y(x_0) = 0$ and I found that $y'(x_0) \in \{0,b\}$. So one solution is $y(x) \sim 0 + 0\times (x-x_0) + \cdots$ while the other is $y(x) \sim 0 + b(x-x_0) + \cdots$. Does this look familiar? These correspond to $y \equiv 0$ and $y = bx - bx_0$ which are the two solutions you get when you put $a=0.$2012-08-31
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    @FlybyNight This equation comes from a physics problem. $y$ represents the particle's speed and $x$ is its position. $-\frac{a}{x}$ represents the force of attraction. $by$ is a resistance force. At inital moment the particle is at rest at distance $x_0$ from the origin. That means $y(x_0)=0$. Thus, it follows from the physical considerations that the solution with $a>0$ exists. What exactly happens at inital moment mathematically does not matter.2012-09-01
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    @S4M $y'(x)$ is acceleration. There is no the acceleration of particle barrier like the speed of particle barrier.2012-09-01
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    @MartinGales: Like I said: you've copied something wrong down. The speed of the particle is given by the rate of change of position. Position is given by $y$ while speed is given by $y'$. Your boundary condition should be $y'(x_0) = 0$ and not $y(x_0) = 0.$2012-09-02
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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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    Thank you. If y(x0)=0 then all OK?2012-08-31
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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