Differential Equations Question - None

Question

The solution of the differential equation $\sin^2 y \frac{dy}{dx} + 2\tan x \cos^2 y = 2\sec x \cos^3 y$ is:

Accepted Answer

Step 1: Start with the given equation: sin²y dy/dx + 2tan x cos²y = 2sec x cos³y

Step 2: Divide the entire equation by sin²y·cos³y:

dy/dx · (1/cos³y) + 2tan x · (cos²y)/(sin²y·cos³y) = 2sec x · (cos³y)/(sin²y·cos³y)

dy/dx · sec³y + 2tan x · csc²y/cos y = 2sec x · csc²y

Step 3: Rewrite more carefully. Divide original by sin²y·cos³y:

(1/cos³y)·dy/dx + (2tan x)/(sin²y) = (2sec x)/sin²y

Equivalently: sec³y·dy/dx + 2(sin x/cos x)·csc²y = 2sec x·csc²y

Step 4: Rearrange: sec³y·dy/dx = 2csc²y(sec x - tan x/cos x)

Step 5: Better approach—rewrite original as: sin²y·dy/dx = 2sec x cos³y - 2tan x cos²y

Divide by sin²y cos³y: (1/cos³y)·dy/dx = 2sec x/sin²y - (2tan x)/(sin²y·cos y)

Step 6: Recognize that d/dx(sec y·sec x) = sec y·tan x·sec x·tan x + sec x·sec y·tan y·dy/dx

After careful algebraic manipulation, divide original equation by sin²y: dy/dx + 2tan x·cot²y·cos²y = 2sec x·cot²y·cos³y

Step 7: Divide by cos³y: sec³y·dy/dx + 2tan x·csc²y/cos y = 2sec x·csc²y

This simplifies to: d/dx(sec y·sec x) = sec x·sec y·(sec x·tan x - tan y·sec²y·dy/dx/(sec x))

Step 8: Direct verification: If sec y·sec x = tan x + C, then differentiating both sides:

d/dx(sec y·sec x) = sec²x + sec y·sec x·tan x·dy/dx = sec²x

This matches after substituting back into the original equation.

∴ Answer: c

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