Consider the differential equation $\frac{dy}{dx}+y\tan x=\sec x$, then which of the following statements are correct? (A) It is homogeneous Choose the correct answer from the options given below: |
(A) and (D) only (B) and (C) only (C) and (D) only (B), (C) and (D) only |
(B) and (C) only |
The correct answer is Option (2) → (B) and (C) only (A) It is homogeneous (Incorrect) Given differential equation: $\frac{dy}{dx} + y \tan x = \sec x$ Check (A): Homogeneous? A differential equation is homogeneous if it can be written as $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$ or all terms are of same degree. Here, RHS is $\sec x$, LHS has $y$ and $\tan x$ — not a function of $\frac{y}{x}$, nor are terms of the same degree. ⇒ ❌ Not homogeneous Check (B): Integrating factor? This is a linear differential equation in standard form: $\frac{dy}{dx} + P(x) y = Q(x)$, with $P(x) = \tan x$, $Q(x) = \sec x$ Integrating factor (I.F.): $e^{\int \tan x \, dx} = e^{-\ln|\cos x|} = \sec x$ ⇒ ✔️ Integrating factor is $\sec x$ Check (C): General solution? Multiply both sides by $\sec x$ (I.F.): $\sec x \cdot \frac{dy}{dx} + y \sec x \tan x = \sec^2 x$ Left side becomes: $\frac{d}{dx}(y \sec x) = \sec^2 x$ Integrate: $y \sec x = \tan x + C$ ⇒ ✔️ General solution is $y \sec x = \tan x + C$ Check (D): Degree? The differential equation is of order 1 and degree 1 (no radicals or powers of derivative) ⇒ ❌ Degree is defined |