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 particular 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 $(1)$:

\(\ds \paren {x^2 + 2 p} \dfrac {\d p} {\d x} + 2 x p\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds 2 x p \rd x + \paren {x^2 + 2 p} \rd p\)

\(=\)

\(\ds 0\)

\(\text {(2)}: \quad\)

\(\ds \leadsto \ \ \)

\(\ds p \paren {x^2 + p}\)

\(=\)

\(\ds C_1\)

Bernoulli's Equation: $2 x y \rd x + \paren {x^2 + 2 y} \rd y = 0$

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:

\(\ds p x^2\)

\(=\)

\(\ds -p^2\)

\(\ds \leadsto \ \ \)

\(\ds p \paren {x^2 - p}\)

\(=\)

\(\ds 0\)

There are two possibilities here:

\(\ds p\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \dfrac {\d y} {\d x}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds C_2\)

From our initial condition:

$y = 1$ when $x = 0$

gives us:

$C_2 = 1$

and so the solution is obtained:

$y = 1$

$\Box$

The other option is:

\(\ds p = \dfrac {\d y} {\d x}\)

\(=\)

\(\ds -x^2\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds -\int x^2 \rd x\)

\(\text {(3)}: \quad\)

\(\ds \)

\(=\)

\(\ds -\frac {x^3} 3 + C_2\)

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.

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.11$: Problem $2 \ \text{(a)}$