Question

The solution of Differential equation dy/dx =secx - y 10x is​

Answer 1

Appropriate Question :-

Solve the Differential equation :-

$\rm :\longmapsto\:\dfrac{dy}{dx} = secx \: - \: y \: tanx$

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

Given differential equation is

$\rm :\longmapsto\:\dfrac{dy}{dx} = secx \: - \: y \: tanx$

can be rewritten as

$\rm :\longmapsto\:\dfrac{dy}{dx} + \: y \: tanx= secx \:$

Its a linear differential equation,

So, on comparing with

$\rm :\longmapsto\:\dfrac{dy}{dx} + \: py= q \: \: where \: p \: and \: q \: \in \: f(x)$

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

and

$\rm :\longmapsto\:q = secx$

Now, Integrating factor is

$\rm :\longmapsto\:IF = {e} \: ^{\displaystyle\int\tt \: p \: dx}$

$\rm :\longmapsto\:IF = {e} \: ^{\displaystyle\int\tt \: tanx \: dx}$

$\rm :\longmapsto\:IF = {e}^{log \: secx}$

$\red{\bigg \{ \because \:\displaystyle\int\tt \: tanx \: dx = log \: secx \: + \: c \bigg \}}$

$\rm :\implies\:IF = secx$

$\red{\bigg \{ \because \: {e}^{logx} = x\bigg \}}$

So,

Solution of Linear Differential equation is

$\rm :\longmapsto\:y \times IF = \displaystyle\int\tt \: (q \times IF) \: dx$

$\rm :\longmapsto\:y \times secx = \displaystyle\int\tt \: secx \times secx \: dx$

$\rm :\longmapsto\:y \: secx = \displaystyle\int\tt \: {sec}^{2}x \: dx$

$\rm :\longmapsto\:y \: secx = tanx + c$

Hence,

Solution of differential equation

$\bf :\longmapsto\:\dfrac{dy}{dx} = secx \: - \: y \: tanx$

is

$\bf :\longmapsto\:y \: secx = tanx + c$