MHTCET 20 Apr 2026 Shift 1 Paper

Concept:For a linear differential equation $\frac{dy}{dx} + Py = Q$, the integrating factor is $\text{I.F.} = e^{\int P \, dx}$.Given that $\tan x$ is the integrating factor, we can determine $P$ by comparing and differentiating.Explanation:The integrating factor is given as $\tan x$, so we write:$$e^{\int P \, dx} = \tan x$$Taking natural logarithm on both sides:$$\int P \, dx = \log(\tan x)$$Differentiate both sides with respect to $x$:$$P = \frac{d}{dx}\left[\log(\tan x)\right]$$Using the derivative rule $\frac{d}{dx}\log f(x) = \frac{f'(x)}{f(x)}$, we get:$$P = \frac{\sec^2 x}{\tan x}$$Simplify step by step:$$P = \frac{1}{\cos^2 x} \times \frac{\cos x}{\sin x}$$$$P = \frac{1}{\sin x \cos x}$$Using the identity $\sin 2x = 2 \sin x \cos x$, we obtain:$$P = \frac{2}{\sin 2x} = 2\csc 2x$$Thus, the correct value of $P$ is $2\csc 2x$.Answer:Option D: $2 \csc 2x$