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

\(\displaystyle y''\) \(=\) \(\displaystyle \dfrac 1 {1 - x^2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int \dfrac {\d^2 y} {\d x^2} \rd x\) \(=\) \(\displaystyle \int \frac 1 {1 - x^2} \rd x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \dfrac {\d y} {\d x}\) \(=\) \(\displaystyle \tanh^{-1} x + C\) Primitive of $\dfrac 1 {a^2 - x^2}$: Inverse Hyperbolic Tangent Form
\(\displaystyle \leadsto \ \ \) \(\displaystyle \int \dfrac {\d y} {\d x} \rd x\) \(=\) \(\displaystyle \int \paren {\tanh^{-1} x + C} \rd x\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle y\) \(=\) \(\displaystyle x \tanh^{-1} x + \map \ln {1 - x^2} + C x + D\) Primitive of $\tanh^{-1} x$

$\blacksquare$


Sources