Second Order ODE/x y'' = y' + (y')^3
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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 $1 \ \text{(b)}$