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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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    For clarity: by $x(0)=x_{0}$ do you mean that $x$ is a function of some other variable, say $t$, and $y = y(x(t))$?2012-03-26

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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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    This is no way to solve a differential equation. If $y=d\,e^x$, it is impossible that $y^2+y\,x=c\,e^x$.2012-03-26
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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.