Question

Solve the following differential equation:

dr + (2r cotθ + sin2θ)dθ = 0

Answer 1

dr + (2r cotθ + sin2θ)dθ = 0

∴ `(dr)/(dθ) + (2r cotθ + sin2θ) = 0`

∴ `(dr)/(dθ) + (2 cotθ)r = - sin 2θ`      ...(1)

This is the linear differential equation of the form

`(dr)/(dθ) + P * r = Q`, where P = 2 cotθ and Q = −sin2θ

∴ I.F. = `e^(int Pdθ) = e^(int e cotθ) dθ`

`= e^(2 int cotθ  dθ) = e^(2 log sinθ)`

`= e^(log (sin^2θ)) = sin^2θ`

∴ The solution of (1) is given by

`r * (I.F.) = int * (I.F.) dθ + c`

∴ `r * sin^2θ = int - sin 2θ * sin^2θ  dθ + c`

∴ `r * sin^2θ = int - 2 sinθ cosθ * sin^2θ  dθ + c`

∴ `r * sin^2θ = - 2 int sin^3θ cosθ  dθ + "c"`

Put sin θ = t

∴ cosθ dθ = dt

∴ `r * sin^2θ = - 2 int t^3 dt + c`

∴ `r * sin^2θ = - 2 * t^4/4 + c`

∴ `r * sin^2θ = - 1/2 sin^4θ + c`

∴ `r * sin^2θ + (sin^4θ)/2 = c`

This is the general solution.