# Linear Second Order ODE/y'' = 1 over 1 - x^2

## 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 $\tanh^{-1} x$

$\blacksquare$