# Cos 2 X D Y D X + Y = Tan X - Mathematics

Sum

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

#### Solution

We have,

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

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

$\text{Comparing with }\frac{dy}{dx} + Px = Q,\text{ we get}$

$P = \sec^2 x$

$Q = \left( \tan x \right)\left( \sec^2 x \right)$

Now,

$I . F . = e^{\int \sec^2 x dx} = e^{\tan x}$

So, the solution is given by

$y \times e^{\tan x} = \int\left( \tan x \right)\left( \sec^2 x \right) \times e^{\tan x} dx + C$

$\Rightarrow y e^{\tan x} = I + C . . . . . . . . . . \left( 1 \right)$

Now,

$I = \int\left( \tan x \right)\left( \sec^2 x \right) \times e^{\tan x} dx$

Putting t = tan x, we get

$dt = \sec^2 x dx$

$= t \times \int e^t dt - \int\left( \frac{d t}{d t} \times \int e^t dt \right)dt$

$= t e^t - \int e^t dt$

$= t e^t - e^t$

$\Rightarrow I = \tan x e^{\tan x} - e^{\tan x} = e^{\tan x} \left( \tan x - 1 \right)$

Putting the value of I in (1), we get

$y e^{\tan x} = e^{\tan x} \left( \tan x - 1 \right) + C$

Is there an error in this question or solution?

#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 55 | Page 146