Question

$co{s}^{2}x\frac{dy}{dx}+y=tanx$ $(0\le x<\frac{\pi }{2})$

Answer 1

${\cos }^{2}x\frac{dy}{dx}+y=\tan x$

$\Rightarrow \frac{dy}{dx}+\frac{1}{{\cos }^{2}x}y=\frac{\tan x}{{\cos }^{2}x}$

$\Rightarrow \frac{dy}{dx}{\sec }^{2}xy=\tan x{\sec }^{2}x$

$\therefore \frac{dy}{dx}+p.y=Q\left(x\right)$

$\therefore$ Integrating factor $=e^{\int Pdx}=e^{\int {\sec }^{2}xdx}=e^{tanx}$

$\Rightarrow e^{\int Pdx}.y=\int e^{\int Pdx}Qdx+c$

But $e^{\int Pdx}=e^{\tan x}$

$\Rightarrow e^{\tan x}.y=\int e^{\tan x}.\tan x{\sec }^{2}xdx+c$

Let $\tan x=t$

$\Rightarrow \frac{d}{dx}\tan x=\frac{dt}{dx}$

$\Rightarrow se{c}^{2}x=\frac{dt}{dx}$

$\therefore \int e^{\tan x}\tan xse{c}^{2}xdx$

$\Rightarrow \int e^t.tdt$

$\Rightarrow \int t.e^{t}dt$

$\Rightarrow t{e}^t-e^t+c$ ( integration by parts )

$\therefore e^{\tan x}.y=\tan x{e}^{\tan x}-e^{\tan x}+c$

$\therefore y=\tan x-1+\frac{c}{\tan x}.$

Hence, the answer is $\tan x-1+\frac{c}{\tan x}.$