Second Order ODE/y y'' = y^2 y' + (y')^2

Theorem
The second order ODE:
 * $(1): \quad y y'' = y^2 y' + \left({y'}\right)^2$

subject to the initial conditions:
 * $y = -\dfrac 1 2$ and $y' = 1$ when $x = 0$

has the solution:
 * $2 y - 3 = 8 y \exp \left({\dfrac {3 x} 2}\right)$

Proof
Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:

$(2)$ is a linear first order ODE in the form:
 * $\dfrac {\mathrm d p} {\mathrm d y} + P \left({y}\right) p = Q \left({y}\right)$

where $P \left({y}\right) = -\dfrac 1 y$.

Thus:

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:
 * $\dfrac {\mathrm d} {\mathrm d y} \left({\dfrac p y}\right) = -\dfrac 1 y y = -1$

and the general solution so far is:
 * $(4): \quad \dfrac p y = - y + C_1$

So:

Now to consider the initial conditions:
 * $y = -\dfrac 1 2$ and $y' = 1$ when $x = 0$

After algebra:

When $x = 0$ we have $y = -1/2$:

Differentiating to get $y'$:


 * $y' \left({1 - 2 C_1}\right) = \left({y + C_1}\right) C_1 e^{C_1 x} + e^{C_1 x} y'$

Putting $y' = 1$ when $x = 0$ we get:

So: