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 $\dfrac x {x^2 + a^2}$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \ln y\) | \(=\) | \(\ds \ln \frac {C_1} {\sqrt {1 + x^2} }\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds y\) | \(=\) | \(\ds \frac C {\sqrt {1 + x^2} }\) |
$\blacksquare$