Question

Integrating factor of
dy/dx + y cot x - cosec x = 0 ​

Answer 1

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Answer 2

$Given \: \frac{dy}{dx} + y cot x - cosec x = 0$

$\implies \frac{dy}{dx} + y cot x = cosec x$

$\blue{ It \: is \: in \: the \: form \: of }$

$\blue{ \frac{dy}{dx} + Py = Q }$

It is a linear differential equation in 'y' .

Hence, P = cot x and Q = cosec x

$\red{ Integral \:Factor }$

$= e^{\int P dx } \\= e^{\int cot x \:dx }\\= e^{log|sin x | }$

$\blue{ \because \int cot x \:dx = log | sin x | + c }$

$\green { = sin x + c }$

The general solution is

$y( I.F) \\= \int Q (I.F) \: dx + c$

$\implies y sin x = \int Cosec x sin x \:dx \\= \int \frac{1}{\cancel { sin x } }\times \cancel { sin x } \: dx \\= \int 1 \: dx$

$\green { y sin x = x + c }$

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