Second Order ODE/(x^2 + 2 y') y'' + 2 x y' = 0

Theorem
The second order ODE:
 * $(1): \quad \paren {x^2 + 2 y'} y'' + 2 x y' = 0$

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

has the solution:
 * $y = 1$

or:
 * $3 y + x^3 = 3$

Proof
The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.

Substitute $p$ for $y'$ in $(1)$:

Consider the initial condition:
 * $y' = p = 0$ when $x = 0$

Hence putting $p = x = 0$ in $(2)$ we get:
 * $0 \cdot 0^2 + 0^2 = C_1$


 * $C_1 = 0$

and so $(2)$ becomes:

There are two possibilities here:

From our initial condition:
 * $y = 1$ when $x = 0$

gives us:
 * $C_2 = 1$

and so the solution is obtained:
 * $y = 1$

The other option is:

From our initial condition:
 * $y = 1$ when $x = 0$

Hence putting $x = 0$ and $y = 1$ in $(3)$ we get:
 * $1 = - \dfrac {0^3} 3 = C_2$

and so $C_2 = 1$.

Thus we have:
 * $y + \dfrac {x^3} 3 = 1$

or:
 * $3 y + x^3 = 3$

Hence the result.