Question

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

• A. $y\sec^{3}x = \sec^{2}x + c$
• B. $y\sec^{2}x = \sec x + c$
• C. $y\sin x = \tan x + c$
• D. None of these

Answer 1

Answer: (B)

$y\sec^{2}x = \sec x + c$

Answer 2

$\frac{dy}{dx}$ +2y tan x = sin x is a linear differential equation of then form $\frac{dy}{dx} + yƒ(x) = g(x)$

$\therefore$ I.F. =

$e^{\int_{}^{}{ƒ(x)dx}} = e^{\int_{}^{}{2\tan xdx}} = e^{2\log(\sec x)} = e^{{logsec}^{2}x} = \sec^{2}x$Hence, the solution is y(I.F.) = $\int_{}^{}{g(x)}I.F.dx + c$

$y(\sec^{2}x) = \int_{}^{}{\sin x\sec^{2}xdx + c}$

$\Rightarrow y\sec^{2}x = \int_{}^{}{\sec x\tan xdx + c \Rightarrow y\sec^{2}}x = \sec x + c$