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Solve the following.
Solve: sin x (dy)/(dx) - y = sin x tanx/2.
Solve: $\sin x \frac{dy}{dx} - y = \sin x \tan\frac{x}{2}$.
Answer
Answer
Divide by $\sin x$: $\frac{dy}{dx} - \frac{1}{\sin x}y = \tan\frac{x}{2}$, a linear equation.
Integrating factor: $\text{I.F.} = e^{-\int \csc x\,dx} = e^{-\log\tan\frac x2} = \cot\frac{x}{2}$.
$y\cot\frac{x}{2} = \int \tan\frac{x}{2}\cot\frac{x}{2}\,dx = \int 1\,dx = x + C$
Hence $y = (x + C)\tan\frac{x}{2}$.
Final answer: $y = (x + C)\tan\frac{x}{2}$