Solve the second-order equation $y''=(2y+3)(y')^2$ Solve the second-order equation $y''=(2y+3)(y')^2$
My Trial
Let $z=z(y):\;y'=z.$ Then,
\begin{align} y''=\dfrac{dz}{dx}=\dfrac{dz}{dy} \dfrac{dy}{dx}=z'\dfrac{dy}{dx}  .\end{align} 
So,
\begin{align} \dfrac{dz}{dx}=(2y+3)(z)^2\iff \dfrac{1}{z^2}\dfrac{dz}{dx}=(2y+3)\iff \dfrac{1}{z^2}dz=(2y+3)dx\end{align} 
I'm stuck here as I am not sure of how to deal with the right-hand side, maybe I should treat it as partial integration or not. Can anyone provide a hint or help me continue?
 A: If you write
$$\frac{(y')'}{y'}=(2y+3)y'$$ you can immediately integrate
$$\log y'=y^2+3y+C,$$
or
$$y'=Ce^{y^2+3y}.$$
This is a separable first-order equation, but with no closed-form solution.
A: You are getting
$$y''=z'\frac{dy}{dx}=zz'$$
(because $\frac{dy}{dx}=z$ by assumption)
Now, using the correct substitution, we get, 
$$\frac{dz}{dy}=(2y+3)z$$
which is a variable separable form.
Hope it help:)
A: Hint
You can make the problem simpler if you remember that
$$\frac{d^2y}{dx^2}=-\frac{ {\frac{d^2x}{dy^2}}} {\left(\frac{dx}{dy}\right)^3}$$ This would give
$$x''+(2 y+3) x'=0$$ Now, $p=x'$ which make things simple (separable), then $p$ and, at the price of a small change of variable, a very standard integral.
A: $$y''=(2y+3)(y')^2$$
$$\frac{y''}{y'}=(2y+3)y'$$
$$\ln(y')=y^2+3y+\text{constant}$$
$$\frac{dy}{dx}=y'=c_1\exp(y^2+3y)$$
$$x=\frac{1}{c_1}\int\exp\big(-(y^2+3y)\big)dy$$
The integral cannot be expressed with a finite number of elementary functions. A closed form requires a special function, namely the erf function.
$$x=\frac{1}{c_1} \left(\text{erf}\left(y+\frac32\right)+c_2\right)$$
Thus the final result is :
$$\boxed{y(x)=-\frac23+\text{erf}^{-1}(c_1x-c_2)}$$
The erf$^{-1}(x)$ function is a standard function which is implemented in the symbolic math softwares. 
http://mathworld.wolfram.com/InverseErf.html
