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प्रश्न
Solve the following differential equation dr + (2r cot θ + sin 2θ) dθ = 0.
उत्तर
dr + (2r cot θ + sin 2θ) dθ = 0
∴ `"dr"/("d" theta) + (2"r" cot theta + sin 2 theta) = 0`
∴ `"dr"/("d" theta) + (2 cot theta)"r" = - sin 2 theta` ...(1)
This is the linear differential equation of the form
`"dr"/("d" theta) + "P" * "r" = "Q"`, where P = 2 cot θ and Q = - sin 2θ
∴ I.F. = `"e"^(int "P d" theta) = "e"^(int "e" cot theta) "d"theta`
`= "e"^(2 int cot theta "d" theta) = "e"^(2 log sin theta)`
`= "e"^(log (sin^2 theta)) = sin^2 theta`
∴ the solution of (1) is given by
`"r" * ("I.F.") = int * ("I.F.") "d"theta + "c"`
∴ `"r" * sin^2theta = int - sin 2 theta * sin^2 theta "d" theta + "c"`
∴ `"r" * sin^2theta = int - 2 sin theta cos theta * sin^2theta "d" theta + "c"`
∴ `"r" * sin^2theta = - 2 int sin^3 theta cos theta "d" theta + "c"`
Put sin θ = t
∴ cos θ dθ = dt
∴ `"r" * sin^2theta = - 2 int "t"^3 "dt" + "c"`
∴ `"r" * sin^2theta = - 2 * "t"^4/4 + "c"`
∴ `"r" * sin^2theta = - 1/2 sin^4 theta + "c"`
∴ `"r" * sin^2theta + ("sin"^4 theta)/2 = "c"`
This is the general solution.
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