Consider the differential equation dy/dx+ytan x=sec x, then which of…

The correct answer is Option (2) → (B) and (C) only

(A) It is homogeneous (Incorrect)
(B) It has $\sec x$ as its integrating factor (Correct)
(C) It's general solution is $y \sec x = \tan x + c$, where c is arbitary constant. (Correct)
(D) It's degree is not defined (Incorrect)

Given differential equation:

$\frac{dy}{dx} + y \tan x = \sec x$

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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

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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 $\sec x$

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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 $y \sec x = \tan x + C$

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Check (D): Degree?

The differential equation is of order 1 and degree 1 (no radicals or powers of derivative)

⇒ ❌ Degree is defined