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
$e^{\int Pdx}=e^{\int 2\tan xdx}$
$=e^{2\log \sec x}=e^{\log {\sec }^{2}x}$
$\therefore I.F.={\sec }^{2}x$
Now solution is,
y . (I.F.) = $\int$ I.F. $\times$ Q + c
$\therefore y.{\sec }^{2}x=\int {\sec }^{2}x.\sin xdx+c$
$\Rightarrow y.{\sec }^{2}x=\int \sec x.\tan xdx+c$
$\Rightarrow y{\sec }^{2}x=\sec x+c$