Second Order ODE/x y'' = y' + (y')^3

Theorem

The second order ODE:

$(1): \quad x y = y' + \paren {y'}^3$

has the general solution:

$x^2 + \paren {y - C_2}^2 = C_1^2$

Proof

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

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

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

\(=\)

\(\ds p + p^3\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \frac x {\sqrt {C_1^2 - x^2} }\)

First Order ODE: $x \rd y = \paren {y + y^3} \rd x$

\(\ds \leadsto \ \ \)

\(\ds \int \rd y\)

\(=\)

\(\ds \int \frac x {\sqrt {C_1^2 - x^2} }\)

Solution to Separable Differential Equation

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds -\sqrt {C_1^2 - x^2} + C_2\)

Primitive of $\dfrac x {\sqrt{a^2 - x^2} }$

\(\ds \leadsto \ \ \)

\(\ds x^2 + \paren {y - C_2}^2\)

\(=\)

\(\ds C_1^2\)

rearranging

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.11$: Problem $1 \ \text{(b)}$