First Order ODE/x dy = (y + y^3) dx

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Theorem

The first order ODE:

$(1): \quad x \dfrac {\d y} {\d x} = y + y^3$

has the general solution:

$y = \dfrac x {\sqrt {C^2 - x^2} }$


Proof

\(\ds x \dfrac {\d y} {\d x}\) \(=\) \(\ds y + y^3\)
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d y} {y \paren {1 + y^2} }\) \(=\) \(\ds \int \frac {\d x} x\) Solution to Separable Differential Equation
\(\ds \leadsto \ \ \) \(\ds \frac 1 2 \, \map \ln {\frac {y^2} {y^2 + 1} }\) \(=\) \(\ds \ln x + \ln k\) Primitive of $\dfrac 1 {x \paren {x^2 + a^2} }$
\(\ds \) \(=\) \(\ds \map \ln {k x}\)
\(\ds \leadsto \ \ \) \(\ds \map \ln {\frac {y^2} {y^2 + 1} }\) \(=\) \(\ds \map \ln {k^2 x^2}\)
\(\ds \leadsto \ \ \) \(\ds \frac {y^2} {y^2 + 1}\) \(=\) \(\ds k^2 x^2\)
\(\ds \leadsto \ \ \) \(\ds \frac {y^2 + 1} {y^2}\) \(=\) \(\ds \frac 1 {k^2 x^2}\)
\(\ds \leadsto \ \ \) \(\ds 1 + \frac 1 {y^2}\) \(=\) \(\ds \frac 1 {k^2 x^2}\)
\(\ds \leadsto \ \ \) \(\ds \frac 1 {y^2}\) \(=\) \(\ds \frac 1 {k^2 x^2} - 1\)
\(\ds \) \(=\) \(\ds \frac {1 - k^2 x^2} {k^2 x^2}\)
\(\ds \leadsto \ \ \) \(\ds y^2\) \(=\) \(\ds \frac {k^2 x^2} {1 - k^2 x^2}\)
\(\ds \leadsto \ \ \) \(\ds y^2\) \(=\) \(\ds \frac {x^2} {\frac 1 {k^2} - x^2}\)
\(\ds \leadsto \ \ \) \(\ds y^2\) \(=\) \(\ds \frac {x^2} {C^2 - x^2}\) letting $C = \dfrac 1 k$
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \frac x {\sqrt {C^2 - x^2} }\)

$\blacksquare$