Linear Second Order ODE/y'' = 1 over 1 - x^2
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Theorem
The second order ODE:
- $(1): \quad y = \dfrac 1 {1 - x^2}$
has the general solution:
- $y = x \tanh^{-1} x + \map \ln {1 - x^2} + C x + D$
Proof
| \(\ds y\) | \(=\) | \(\ds \dfrac 1 {1 - x^2}\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \int \dfrac {\d^2 y} {\d x^2} \rd x\) | \(=\) | \(\ds \int \frac 1 {1 - x^2} \rd x\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \dfrac {\d y} {\d x}\) | \(=\) | \(\ds \tanh^{-1} x + C\) | Primitive of $\dfrac 1 {a^2 - x^2}$: Inverse Hyperbolic Tangent Form | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \int \dfrac {\d y} {\d x} \rd x\) | \(=\) | \(\ds \int \paren {\tanh^{-1} x + C} \rd x\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds x \tanh^{-1} x + \map \ln {1 - x^2} + C x + D\) | Primitive of $\tanh^{-1} x$ |
$\blacksquare$
Sources
- 1963: Morris Tenenbaum and Harry Pollard: Ordinary Differential Equations ... (previous) ... (next): Chapter $1$: Basic Concepts: Lesson $3$: The Differential Equation: $(3.13)$