Question

Find the particular solution of the differential equation $e^{x}tanydx+(2-e^x)se{c}^{2}ydy=0$, given that $y=\frac{\pi }{4}$ when x = 0
OR
Find the particular solution of the differential equation $\frac{dy}{dx}+2ytanx=sinx$, given that y = 0 when $x=\frac{\pi }{3}$.

Answer 1

We have $e^{x}tanydx+(2-e^x)se{c}^{2}ydy=0\Rightarrow \int \frac{e^{x}dx}{2-e^x}+\int \frac{se{c}^{2}ydy}{tany}=0$

Put $e^x=t\Rightarrow e^{x}dx=dt$ in 1st integral.

and , tan y = u $\Rightarrow se{c}^{2}ydy=du$ in 2nd integral.

$\therefore \int \frac{dt}{2-t}+\int \frac{du}{u}=0\Rightarrow log|u|-log|2-t|=log|c|log\left|\frac{tany}{2-e^x}\right|=log|c|\Rightarrow \frac{tany}{2-e^x}=c$

Given that $y=\frac{\pi }{4}$ when x = 0 so, $\frac{tan\frac{\pi }{4}}{2-e^0}=c$ $\Rightarrow c=1$

Hence required solution is , $tany=2-e^x$ or $y=ta{n}^{-1}(2-e^x)$

OR We have $\frac{dy}{dx}+2ytanx=sinX$

$\Rightarrow \frac{dy}{dx}+(2+tanx)y=sinx$

This differential equation is of the form $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right),$ where P(x) = 2 tan x, Q(x) = sin x.

So, integrating factor, $I.F=e^{\int 2tanxdx}=e^{2logsecx}=se{c}^{2}x$

So, the solution is : $yse{c}^{2}x=\int se{c}^{2}xsinxdx+c\Rightarrow yse{c}^{2}x=\int secxtanxdx+c=secx+c$

Given that y = 0 when $x=\frac{\pi }{3}$ so, $0\times se{c}^2\frac{\pi }{3}=sec\frac{\pi }{3}+c\Rightarrow c=-2$

Hence the required solution is $yse{c}^{2}x=secx-2$ or $y=cosx-2co{s}^{2}x$