What is the solution of the differential equation (dy-dx)+\cos x(dy+dx)=0?
- A. y=\tan(\frac{x}{2})-x+c
- B. y=\frac{1}{2}\tan(\frac{x}{2})-x+c
- C. y=2\tan(\frac{x}{2})-x+c ✓
- D. y=\tan(\frac{x}{2})-2x+c
Correct Answer: C. y=2\tan(\frac{x}{2})-x+c
Explanation
Rearranging terms gives dy(1+\cos x) - dx(1-\cos x) = 0, which separates into dy = \frac{1-\cos x}{1+\cos x}dx. Using the half-angle identities 1-\cos x = 2\sin^2(x/2) and 1+\cos x = 2\cos^2(x/2), this simplifies to dy = \tan^2(x/2) dx = (\sec^2(x/2) - 1) dx. Integrating yields y = 2\tan(\frac{x}{2}) - x + c.
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