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in Differential Equations by (29.6k points)
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Solve the following differential equation:

dr + (2r cot θ + sin 2θ) dθ = 0

1 Answer

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Best answer

dr + (2r cot θ + sin 2θ) dθ = 0

\(\therefore\) \(\cfrac{dr}{d\theta}\) + (2r cot θ + sin 2θ) = 0

\(\therefore\) \(\cfrac{dr}{d\theta}\) + (2 cot θ)r = -sin 2θ ……(1)

This is the linear differential equation of the form dr

\(\cfrac{dr}{d\theta}\) + P . r = Q, where P = 2 cot θ and Q = -sin 2θ

\(\therefore\) the solution of (1) is given by

This is the general solution.

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