Second Order ODE/y'' = 1 + (y')^2/Proof 2

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Theorem

The second order ODE:

$(1): \quad y'' = 1 + \paren {y'}^2$

has the general solution:

$y = \map {\ln \sec} {x + C_1} + C_2$


Proof

The proof proceeds by using Solution of Second Order Differential Equation with Missing Dependent Variable.

Substitute $p$ for $y'$ in $(1)$:

\(\ds \dfrac {\d p} {\d x}\) \(=\) \(\ds p^2 + 1\) where $p = \dfrac {\d y} {\d x}$
\(\ds \leadsto \ \ \) \(\ds \int \rd x\) \(=\) \(\ds \int \frac {\d p} {p^2 + 1}\) Solution to Separable Differential Equation
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \arctan p + A_1\) Primitive of $\dfrac 1 {x^2 + a^2}$
\(\ds \leadsto \ \ \) \(\ds p = \frac {\d y} {\d x}\) \(=\) \(\ds \map \tan {x - A_1}\)
\(\ds \leadsto \ \ \) \(\ds \int \rd y\) \(=\) \(\ds \int \map \tan {x - A_1} \rd x\) Solution to Separable Differential Equation
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \map {\ln \sec} {x - A_1} + A_2\) Primitive of $\tan a x$
\(\ds \) \(=\) \(\ds \map {\ln \sec} {x + C_1} + C_2\) rearranging constants

$\blacksquare$


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