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

Theorem

The second order ODE:

$(1): \quad y + 2 x \paren {y'}^2 = 0$

has the general solution:

$C_1 \map \arctan {C_1 x} = y + C_2$

Proof

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

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

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

\(=\)

\(\ds -2 x p^2\)

\(\ds \leadsto \ \ \)

\(\ds \int \frac {\d p} {p^2}\)

\(=\)

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

Solution to Separable Differential Equation

\(\ds \leadsto \ \ \)

\(\ds -\frac 1 p\)

\(=\)

\(\ds -x^2 + k^2\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds x^2 + k^2\)

substituting back for $p$

\(\ds \leadsto \ \ \)

\(\ds \int \frac {\d x} {x^2 + k^2}\)

\(=\)

\(\ds \int \rd y\)

Solution to Separable Differential Equation

\(\ds \leadsto \ \ \)

\(\ds \frac 1 k \map \arctan {\frac x k}\)

\(=\)

\(\ds y + C_2\)

Primitive of $\dfrac 1 {x^2 + a^2}$

\(\ds \leadsto \ \ \)

\(\ds C_1 \map \arctan {C_1 x}\)

\(=\)

\(\ds y + C_2\)

setting $C_1 = \dfrac 1 k$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $11$