A Second order ODE 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.
 A: 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.
A: 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 :
$$y=c\ \frac{\exp(x)}{y+x}\quad \Rightarrow\quad y^2+yx=c\ \exp(x).$$ 
