Particular solution of the Differential Equation given

Particular solution of the Differential Equation given:

(2a^2-r^2)dr = r^3sintheta dtheta; when theta = 0, r=a.

Please help! Thanks!

2∙a² - r² dr = r³∙sin(θ) dθ
<=>
2∙(a²/r³) - (1/r) dr = sin(θ) dθ
=>
∫ 2∙(a²/r³) - (1/r) dr = ∫ sin(θ) dθ
=>
-(a²/r²) - ln(r) = -cos(θ) + C
So the general solution for this differential equation is given by
(a²/r²) + ln(r) = cos(θ) - C

Apply initial condition to evaluate constant C
(a²/a²) + ln(a) = cos(0) - C
<=>
1 + ln(a) = 1 - C
<=>
C = - ln(a)

So the particular solution to this initial value problem is:
(a²/r²) + ln(r) = cos(θ) + ln(a)
<=>
(a/r)² + ln(r) - ln(a) = cos(θ)
<=>
(a/r)² + ln(r/a) = cos(θ)

data insufficient i guess

Respect to r it is -2/a^2