Linear Second Order ODE/(1 + x^2) y'' + x y' = 0

Theorem

The second order ODE:

$\paren {1 + x^2} y + x y' = 0$

has the general solution:

$y = C_1 \map \ln {x + \sqrt {x^2 + 1} } + C_2$

Proof

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

Substitute $p$ for $y'$:

\(\ds \paren {1 + x^2} \dfrac {\d p} {\d x} + x p\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds p = \dfrac {\d y} {\d x}\)

\(=\)

\(\ds \frac {C_1} {\sqrt {1 + x^2} }\)

First Order ODE: $\paren {1 + x^2} y' + x y = 0$

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds C_1 \int \frac 1 {\sqrt {1 + x^2} } \rd x\)

\(\ds \leadsto \ \ \)

\(\ds y\)

\(=\)

\(\ds C_1 \map \ln {x + \sqrt {x^2 + 1} } + C_2\)

Primitive of $\dfrac 1 {\sqrt{x^2 + a^2} }$

$\blacksquare$

Sources

• 1972: George F. Simmons: Differential Equations ... (previous) ... (next): Miscellaneous Problems for Chapter $2$: Problem $23$