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What might be a solution to the differential equation of the form $$xy''=c{y\over y+d}$$ where $y=y(x)$ and $c,d $ are constants? I am supposed to simply "state" a solution to this, but I don;t think it is all that obvious.

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Not obvious at all, unless you're allowed to use the trivial solution $y=0$. Are there any boundary conditions? –  in_wolframAlpha_we_trust Mar 26 '12 at 11:06
    
Well, $y(x)=0$ is a solution of your ODE. –  Pacciu Mar 26 '12 at 11:07
    
@Pacciu: Yes, I saw that too :) But I don't think that is what they are after...! –  integral Mar 26 '12 at 11:19
    
@in_wolfram_we_trust: (Nice username name!) The boundary conditions would be $y'(0)=0$, and $x(0)=x_0$. Sorry for omitting them earlier, I just thought I could plug them in after "spotting" the general solution. –  integral Mar 26 '12 at 11:20
    
For clarity: by $x(0)=x_{0}$ do you mean that $x$ is a function of some other variable, say $t$, and $y = y(x(t))$? –  in_wolframAlpha_we_trust Mar 26 '12 at 12:21

2 Answers 2

The only thing I can think of is to try to obtain a solution as a power series. First of all, the change of variable $y=d\,z$ changes the equation into $$ x\,z''=\frac{\alpha\,z}{z+1},\quad\alpha=\frac{c}{d}. $$ Look for a solution of the form $$ z(x)=\sum_{n=0}^\infty a_nx^n=a_0+a_1x+a_2x^2+a_3x^3+\dots $$ It is easy tio check that the only possible value for $a_0$ is $a_0=0$. Then $$\begin{align*} x\,z''&=2\,a_2x+6\,a_3x^2+12\,a_4x^3+\dots\\ \frac{\alpha\,z}{z+1}&=\alpha\,a_1x+\alpha(a_2-a_1^2)x^2+\alpha(a_1^3-2\,a_1a_2+a_3)x^3+\dots \end{align*}$$ Equating coefficients of equal powers, we can find an expression for $a_n$ in terms of $a_1,\dots,a_{n-1}$: $$\begin{align*} a_2&=\frac{\alpha\,a_1}{2},\\ a_3&=\frac{\alpha}{6}(a_2-a_1^2),\\ a_4&=\frac{\alpha}{12}(a_1^3 - 2\,a_1 a_2 + a_3),\\ \dots&=\dots \end{align*}$$ The value of $a_1=z'(0)$ is unrestricted. Of course, to prove that $z$ is in fact a solution, one must show that the series has a positive radius of convergence.

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Thank you, Julian. –  integral Mar 26 '12 at 18:58

Try $y=d\, \exp(x)\, (d\neq 0)$. Then we have

$$x (d\, \exp(x))=c \frac{d\, \exp(x)}{d \,\exp(x)+d}$$ which after simplification gives an implicit solution:

$$y=c\ \frac{\exp(x)}{x(\exp(x) +1)}.$$ Which can be further simplied as :

$$y=c\ \frac{\exp(x)}{y+x}\quad \Rightarrow\quad y^2+yx=c\ \exp(x).$$

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Where's the solution? The expression is explicit. –  Pedro Tamaroff Mar 26 '12 at 16:51
    
Thank you, Sri Pot. –  integral Mar 26 '12 at 18:59
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This is no way to solve a differential equation. If $y=d\,e^x$, it is impossible that $y^2+y\,x=c\,e^x$. –  Julián Aguirre Mar 26 '12 at 19:55

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