Question

tan y dy/dx+tanx=cosycos2x

Answer 1

$\large\underline{\sf{Correct \:Question - }}$

$\sf \: Solve : \: tany\dfrac{dy}{dx} + tanx = cosy \: {cos}^{2} x$

$\large\underline{\bold{Solution-}}$

The given Differential equation is

$\sf \: \: tany\dfrac{dy}{dx} + tanx = cosy \: {cos}^{2} x$

$\rm :\longmapsto\: \bf \: Divide \: whole \: equation \: by \: cosy$

$\rm \:\: secy \: tany\dfrac{dy}{dx} + secy \: tanx = \: {cos}^{2} x - - (1)$

$\sf \: Put \: secy \: = t\rm :\implies\:secy \: tany \: \dfrac{dy}{dx} \: = \dfrac{dt}{dx}$

So, equation (1) can reduced to

$\sf \: \dfrac{dt}{dx} + t \: tanx \: = {cos}^{2} x - - - (2)$

This is linear differential equation,

$\bf \: On \: comparing \: with \: \dfrac{dt}{dx} + pt \: = q$

we get

$\rm :\longmapsto\:p \: = \: tanx$

and

$\rm :\longmapsto\:q \: = \: {cos}^{2} x$

So, Integrating Factor is given by

$\rm \: I.F. \: = \: {e}^{ \int \:p \: dx}$

$\rm \: I.F. \: = \: {e}^{ \int \:tanx \: dx}$

$\rm \: I.F. \: = \: {e}^{logsecx} \: \: \: \: \: (\: \because \: \int \: tanxdx \: = \: logsecx \: )$

$\bf\implies \:I.F. \: = \: secx \: \: \: \: \: \: \: (\because \: {e}^{logx} = x)$

Hence, Solution of equation (2) is given by

$\bf \: t \times I.F. \: = \int \: (q \times I.F. )dx$

On substituting the values, we get

$\rm \: t \: secx = \int( {cos}^{2} x \times secx)dx$

$\rm \: t \: secx = \int \: cosxdx$

$\rm \: t \: secx = sinx \: + \: c$

$\bf\implies \:secy \: secx = sinx + c \: \: \: \: \: \: \: ( \because \: t \: = \: secy)$