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

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

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 $(1)$:

It can be seen that $(2)$ is in the form:
 * $\dfrac {\mathrm d x} {\mathrm d p} + P \left({p}\right) x = Q \left({p}\right) x^n$

where:
 * $P \left({p}\right) = \dfrac 1 {2 p}$
 * $Q \left({p}\right) = -1$
 * $n = -1$

and so is an example of Bernoulli's equation.

By Solution to Bernoulli's Equation it has the general solution:
 * $(3): \quad \displaystyle \frac {\mu \left({p}\right)} {x^{n - 1} } = \left({1 - n}\right) \int Q \left({p}\right) \mu \left({p}\right) \, \mathrm d p + C$

where:
 * $\mu \left({p}\right) = e^{\left({1 - n}\right) \int P \left({p}\right) \, \mathrm d p}$

Thus $\mu \left({x}\right)$ is evaluated:

and so substituting into $(3)$:

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

Hence putting $p = x = 0$ in $(4)$ we get:
 * $0 \cdot 0^2 + 0^2 = C_1$


 * $C_1 = 0$

and so $(4)$ 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 $(5)$ we get:
 * $1 = - \dfrac {0^3} 3 = C_2$

and so $C_2 = 1$.

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

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

Hence the result.