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

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Theorem

The first order ODE:

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

has the general solution:

$y = \dfrac C {\sqrt {1 + x^2} }$


Proof

\(\ds \paren {1 + x^2} \dfrac {\d y} {\d x}\) \(=\) \(\ds -x y\)
\(\ds \leadsto \ \ \) \(\ds \int \frac {\d y} y\) \(=\) \(\ds -\int \frac {x \rd x} {1 + x^2}\) Solution to Separable Differential Equation
\(\ds \leadsto \ \ \) \(\ds \ln y\) \(=\) \(\ds -\frac 1 2 \, \map \ln {1 + x^2} + C\) Primitive of $\dfrac x {x^2 + a^2}$
\(\ds \leadsto \ \ \) \(\ds \ln y\) \(=\) \(\ds \ln \frac {C_1} {\sqrt {1 + x^2} }\)
\(\ds \leadsto \ \ \) \(\ds y\) \(=\) \(\ds \frac C {\sqrt {1 + x^2} }\)

$\blacksquare$