Question:

<p><strong>Step 1:</strong> Rearrange the given equation cos x dy = y(sin x - y) dx</p><p>$$\frac{dy}{dx} = \frac{y(\sin x - y)}{\cos x} = y\tan x - \frac{y^2}{\cos x}$$</p><p><strong>Step 2:</strong> Rewrite in standard Bernoulli form with n = 2:</p><p>$$\frac{dy}{dx} - y\tan x = -\frac{y^2}{\cos x}$$</p><p><strong>Step 3:</strong> Divide by y²:</p><p>$$\frac{1}{y^2}\frac{dy}{dx} - \frac{\tan x}{y} = -\frac{1}{\cos x}$$</p><p><strong>Step 4:</strong> Substitute v = 1/y, so dv/dx = -1/y² · dy/dx:</p><p>$$-\frac{dv}{dx} - v\tan x = -\frac{1}{\cos x}$$</p><p>$$\frac{dv}{dx} + v\tan x = \frac{1}{\cos x}$$</p><p><strong>Step 5:</strong> This is linear in v. Integrating factor: $e^{\int \tan x\, dx} = e^{-\ln|\cos x|} = \sec x$</p><p><strong>Step 6:</strong> Multiply by sec x:</p><p>$$\sec x \frac{dv}{dx} + v\sec x \tan x = \sec^2 x$$</p><p>$$\frac{d}{dx}(v\sec x) = \sec^2 x$$</p><p><strong>Step 7:</strong> Integrate:</p><p>$$v\sec x = \tan x + C$$</p><p><strong>Step 8:</strong> Substitute back v = 1/y:</p><p>$$\frac{\sec x}{y} = \tan x + C$$</p><p>∴ Answer: <strong>A</strong> (The solution is $\frac{1}{y\cos x} = \tan x + C$ or equivalent form)</p>