( Tan 2 X + 2 Tan X + 5 ) D Y D X = 2 ( 1 + Tan X ) Sec 2 X - Mathematics

Sum

$(\tan^2 x + 2\tan x + 5)\frac{dy}{dx} = 2(1+\tan x)\sec^2x$

Solution

We have,

$\left( \tan^2 x + 2 \tan x + 5 \right)\frac{dy}{dx} = 2\left( 1 + \tan x \right) \sec^2 x$

$\Rightarrow dy = \frac{2\left( 1 + \tan x \right) \sec^2 x}{\left( \tan^2 x + 2 \tan x + 5 \right)} dx$

Integrating both sides, we get

$\int dy = \int\frac{2\left( 1 + \tan x \right) \sec^2 x}{\left( \tan^2 x + 2 \tan x + 5 \right)} dx . . . . . . . . \left( 1 \right)$
$\text{Putting }\tan^2 x + 2 \tan x + 5 = t$

$\therefore \left( 2 \tan x se c^2 x + 2se c^2 x \right) dx = dt$

$\Rightarrow 2\left( 1 + \tan x \right) \sec^2 x dx = dt$

Therefore (1) becomes,

$\int dy = \int\frac{1}{t} dt$

$\Rightarrow y = \log \left| t \right| + C$

$\Rightarrow y = \log \left| \tan^2 x + 2 \tan x + 5 \right| + C$

Is there an error in this question or solution?

APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 24 | Page 145