Question

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

• A. $y=t-1+c{e}^{-t}$ where $t=\tan x.$
• B. $y=t+1+c{e}^{-t}$ where $t=\tan x.$
• C. $y=t-1-c{e}^{-t}$ where $t=\tan x.$
• D. None of these.

Answer 1

The correct option is A $y=t-1+c{e}^{-t}$ where $t=\tan x.$

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

$\Rightarrow \frac{dy}{dx}+y{\sec }^{2}x=\tan x{\sec }^{2}x$ ....(1)

Here $P={\sec }^{2}x\Rightarrow \int PdP=\int {\sec }^{2}xdx=\tan x$

$\therefore I.F.=e^{\tan x}$

Multiplying (1) by $I.F.$ we get

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

Integrating both sides, we get

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

Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$

$\therefore y{e}^t=\int t{e}^{t}dt=e^t(t-1)+c$

$\Rightarrow y=t-1+c{e}^{-t}$ where $t=\tan x$