Linear First Order ODE/(1 + x^2) dy + 2 x y dx = cotangent x dx

Theorem

$(1): \quad \paren {1 + x^2} \rd y + 2 x y \rd x = \cot x \rd x$

$y = \dfrac {\map \ln {\sin x} } {1 + x^2} + \dfrac C {1 + x^2}$

$(1)$ can be written as:

$(2): \quad \paren {1 + x^2} \dfrac {\rd y} {\rd x} + 2 x y = \cot x$

We notice straight away that:

$\dfrac {\rd} {\rd x} \paren {1 + x^2} = 2 x$

and so:

$\dfrac {\rd} {\rd x} \paren {1 + x^2} y = \cot x$

Thus the general solution becomes:

$\ds \paren {1 + x^2} y$

$=$

$\ds \int \cot x \rd x$

$\ds $

$=$

$\ds \map \ln {\sin x} + C$

or:

$y = \dfrac {\map \ln {\sin x} } {1 + x^2} + \dfrac C {1 + x^2}$

$\blacksquare$