Question

The solution of the equation, $\frac{dy}{dx} + 2y \tan x = \sin x$, is

• A. $y \, \sec^2 \, x = \sec \, x + c $
• B. $y \, \sec \, x = \tan \, x + c $
• C. $y = \sec \, x + c \sec^2 \, x$
• D. none of these

Answer 1

Answer: (A)

Given, $\frac{dy}{dx} + 2y \tan x = \sin x $ .......(i)
Compare the equation (i) with the general equation $\frac{dy}{dx} + Py = Q $
We find that, P = 2 tan x, Q = sin x
Now, integrating factor, I.F. is
$e^{\int Pdx} = e^{\int 2 \tan xdx }$
$ = e^{2\log\sec x} = e^{\log\sec^2x}$
$\therefore I.F. = \sec^{2} x $
Now solution is,
y . (I.F.) = $\int $ I.F. $\times $ Q + c
$\therefore \, y. \sec^2 x = \int \sec^2 x. \sin x dx + c $
$\Rightarrow \, y .\sec^2 \, x = \int \sec x . \tan x dx + c $
$\Rightarrow \, y \sec^2 x = \sec x + c $