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

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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

Using Solution of Second Order Differential Equation with Missing Independent Variable, $(1)$ can be expressed as:

\(\ds p \frac {\d p} {\d y}\) \(=\) \(\ds p^2 + 1\) where $p = \dfrac {\d y} {\d x}$
\(\ds \leadsto \ \ \) \(\ds \int \rd y\) \(=\) \(\ds \int \frac {p \rd p} {p^2 + 1}\) Solution to Separable Differential Equation
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \frac 1 2 \, \map \ln {p^2 + 1} + k\) Primitive of $\dfrac x {x^2 + a^2}$
\(\ds \leadsto \ \ \) \(\ds p = \frac {\d y} {\d x}\) \(=\) \(\ds \sqrt {A_1^2 e^{2 y} - 1}\) where $A_1^2 = e^{2 k}$
\(\text {(2)}: \quad\) \(\ds \leadsto \ \ \) \(\ds \int \rd x\) \(=\) \(\ds \int \frac {\d y} {\sqrt {A_1^2 e^{2 y} - 1} }\) Solution to Separable Differential Equation


Making the subtitution $u = A_1 e^y$:

$\d u = A_1 e^y \rd y = u \rd y$


Thus $(2)$ transforms into:

\(\ds \int \rd x\) \(=\) \(\ds \int \frac {\d u} {u \sqrt {u^2 - 1} }\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \arcsec u + A_2\) Primitive of $\dfrac 1 {x \sqrt {x^2 - a^2} }$
\(\ds \) \(=\) \(\ds \map \arcsec {A_1 e^y} + A_2\)
\(\ds \leadsto \ \ \) \(\ds \map \sec {x - A_2}\) \(=\) \(\ds A_1 e^y\)
\(\ds \leadsto \ \ \) \(\ds \map {\ln \sec} {x - A_2}\) \(=\) \(\ds \ln A_1 + \ln e^y\)
\(\ds \) \(=\) \(\ds \ln A_1 + y\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \map {\ln \sec} {x - A_2} - \ln A_1\)
\(\ds \) \(=\) \(\ds \map {\ln \sec} {x + C_1} + C_2\) setting $C_1 = -A_2$ and $C_2 = -\ln A_1$

$\blacksquare$


Sources

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