Question

The general solution of the differential equation \[ y + \cos x \left( \frac{dy}{dx} \right) - \cos^2 x = 0 \] is

• A. \((\sec x + \tan x) y = x + \cos x + c\)
• B. \((1 + \cos x) y = (x + c) \cos x - \cos^2 x\)
• C. \((1 + \sin x) y = (x + c) \cos x - \cos^2 x\)
• D. \((\sec x + \tan x) y = x - \sin x + c\)

Hint:

For linear differential equations, rewrite in standard form \(\frac{dy}{dx} + P(x) y = Q(x)\), find the integrating factor, and solve. You can also verify each option by substitution.

Answer 1

The Correct Option is C

Given: \[ y + \cos x \left( \frac{dy}{dx} \right) - \cos^2 x = 0 \Rightarrow \cos x \frac{dy}{dx} = \cos^2 x - y \Rightarrow \frac{dy}{dx} = \frac{\cos^2 x - y}{\cos x} \] Now, rewrite this as: \[ \frac{dy}{dx} + \frac{y}{\cos x} = \cos x \] This is a linear differential equation of the form: \[ \frac{dy}{dx} + P(x) y = Q(x) \] with \(P(x) = \frac{1}{\cos x}\) and \(Q(x) = \cos x\). Now, the integrating factor (I.F.) is: \[ \exp\left(\int \frac{1}{\cos x} dx\right) = \exp\left(\int \sec x \, dx\right) = e^{\ln|\sec x + \tan x|} = |\sec x + \tan x| \] Multiplying through by the I.F.: \[ |\sec x + \tan x| y = \int |\sec x + \tan x| \cdot \cos x \, dx \] This is complex to evaluate directly in this form, so try an alternate method. Let us solve the equation by using the correct answer to verify. 
Try Option (3): Let \[ (1 + \sin x) y = (x + c) \cos x - \cos^2 x \] Differentiate both sides: LHS: \[ \frac{d}{dx} \left( (1 + \sin x) y \right) = \cos x \cdot y + (1 + \sin x) \cdot \frac{dy}{dx} \] RHS: \[ \frac{d}{dx} \left( (x + c) \cos x - \cos^2 x \right) = \cos x - (x + c) \sin x + 2 \cos x \sin x \] Equating both sides and simplifying confirms this is a valid solution.