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

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

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

has the particular solution:
 * $2 y - 3 = 8 y \, \map \exp {\dfrac {3 x} 2}$

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

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 $y'$:


 * $y' \paren {1 - 2 C_1} = \paren {y + C_1} C_1 e^{C_1 x} + e^{C_1 x} y'$

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

So: