Question

The general solution of the differential equation \(\frac{dy}{dx} = y\tan x - y^2\sec x\) is:

• A. \(\tan x = (C + \sec x)y\)
• B. \(\sec y = (C + \tan y)x\)
• C. \(\sec x = (C + \tan x)y\)
• D. \(\tan y = (C + \sec x)x\)

Hint:

Bernoulli equation: substitute \(v = y^{1-n}\) to convert to linear form.

Answer 1

The Correct Option is C

Concept: This is a Bernoulli differential equation of the form \(\frac{dy}{dx} + P(x)y = Q(x)y^n\). Here \(n=2\).

Step 1: Rewrite the equation. \[ \frac{dy}{dx} - y\tan x = -y^2\sec x \]

Step 2: Divide by \(y^2\). \[ y^{-2}\frac{dy}{dx} - y^{-1}\tan x = -\sec x \]

Step 3: Substitute \(v = y^{-1}\). Then \(\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}\). So: \[ -\frac{dv}{dx} - v\tan x = -\sec x \Rightarrow \frac{dv}{dx} + v\tan x = \sec x \]

Step 4: Solve linear DE. Integrating factor \(e^{\int \tan x dx} = e^{\ln|\sec x|} = \sec x\). \[ \frac{d}{dx}(v \sec x) = \sec^2 x \] \[ v \sec x = \tan x + C \] \[ v = \frac{\tan x + C}{\sec x} = \sin x + C\cos x \]

Step 5: Back substitute \(v = 1/y\). \[ \frac{1}{y} = \sin x + C\cos x \Rightarrow y = \frac{1}{\sin x + C\cos x} \] This can be rearranged to \(\sec x = (C + \tan x)y\). Multiply both sides by \(\sec x\): \(1 = (C + \tan x)y \cos x\)? Actually: \[ y = \frac{1}{\sin x + C\cos x} = \frac{1}{\cos x (\tan x + C)} = \frac{\sec x}{\tan x + C} \] \[ \Rightarrow \sec x = (C + \tan x)y \]