Solve the following differential equation:
$\cos^{2} x \frac{dy}{dx} + y = \tan x$

Solution

$\cos^{2} x \frac{dy}{dx} + y = \tan x \Rightarrow \frac{dy}{dx} + \sec^{2} x . y = \sec^{2} x \tan x This equation is in the form of \frac{dy}{dx} + py = Q Here P = \sec^{2} x and Q = \sec^{2} x \tan x Integrating factor, I . F = e^{\int p dx} = e^{\int \sec^{2} x dx} = e^{\tan x}$

The general solution can be given by

y ( I. F ) = $\int$ ( Q x I. F ) dx + C ..........(i)

Let tan x = t

$\frac{d}{dx} (\tan x) = \frac{dt}{dx} \Rightarrow \sec^{2} x = \frac{dt}{dx} \Rightarrow \sec^{2} x dx = dt$

Therefore, equation (i) becomes:

$y e^{\tan x} = \int (e^{t} . t) dt \Rightarrow y e^{\tan x} = \int (e^{t} . t) dt + C \Rightarrow y e^{\tan x} = t . \int e^{t} dt - \int (\frac{d}{dt} (t) . \int e^{t} dt) dt + C \Rightarrow y e^{\tan x} = t . e^{t} - \int e^{t} dt + C \Rightarrow y e^{\tan x} = t . e^{t} - e^{t} + C \Rightarrow y e^{\tan x} = (t - 1) e^{t} + C \Rightarrow y e^{\tan x} = (\tan x - 1) e^{\tan x} + C \Rightarrow y = (\tan x - 1) + C e^{- \tan x}, where C is an arbitary constant .$

For Daily Free Study Material Join wiredfaculty

Differential Equations

Solve the following differential equation:
$\cos^{2} x \frac{dy}{dx} + y = \tan x$

Some More Questions From Differential Equations Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction

For Daily Free Study Material Join wiredfaculty

Differential Equations

Solve the following differential equation:cos2 x dydx + y = tan x

Some More Questions From Differential Equations Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Solve the following differential equation:
$\cos^{2} x \frac{dy}{dx} + y = \tan x$