Advertisement Remove all ads

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

Sum

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

Advertisement Remove all ads

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?
Advertisement Remove all ads

APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 24 | Page 145
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×