Find the particular solution of DE, θ dπ/dθ = 2π, π=9 when θ = -½?

Question

Find the particular solution of DE, θ dπ/dθ = 2π, π=9 when θ = -½

Answer 1

Substituting r for π,
θdr/dθ = 2r
(1/2)dr/r = dθ/θ
(1/2)ln(r) = ln(θ) + ln(C)
ln(r) = 2ln(Cθ)
r = Cθ^2
C = 9 / (1/4) = 36
r = 36θ^2

Answer 2

You can separate variables to get d pi/(2pi)= d(th)/(th). Then you have 1/2* ln (2pi) = ln (th) +C
Since C is a constant, so is 1/ {2lnC}, then,
ln(th) = A ln (2*pi), (th) <0. With the initial condition, -0.69 = A ln 18 or,[ -0.69/ln(18)]=A