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 $C = \dfrac 1 k$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \frac x {\sqrt {C^2 - x^2} }\) |
$\blacksquare$