Second Order ODE/y'' = 1 + (y')^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 1
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 $C_2 = -\ln A_1$ |
$\blacksquare$
Proof 2
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 |
$\blacksquare$
Sources
- 1972: George F. Simmons: Differential Equations ... (previous) ... (next): $\S 2.11$: Problem $3$