The equation is:
$$\frac{1}{1-r} \frac{d}{d r}\bigg((1-r)\frac{d \theta}{d r}\bigg) = -2\times 3.66\times (1-(1-r)^2)\,\theta$$
with boundary conditions: $\theta(r=0) = 0$ and $\frac{d \theta}{d r} = 0$ at $r=1$. To solve this I am employing shooting method and guessing for the value $\frac{d \theta}{d r} = 1.83$ at $r=0$.
I have come up with the following: \begin{aligned} \frac{d \theta}{dr} &= x \, ,\\ \frac{dx}{dr} &= \frac{x}{1-r} - 2\times 3.66\times\theta\, (2r-r^2) \, . \end{aligned} But solving this system of equations does not give me the value zero at $x(r=1)=0$ (ie., $\frac{d \theta}{d r} = 0$ at r=1).
How to split the above equation into two separate first-order ODE's to solve them?
The DE can be analytically solved in terms of confluent hypergeometric functions (below).
http://mathworld.wolfram.com/ConfluentHypergeometricFunctionoftheFirstKind.html
http://mathworld.wolfram.com/ConfluentHypergeometricFunctionoftheSecondKind.html
Of course, this is not the answer expected. But it makes clear the cause of the trouble encountered :
In fact, the initial conditions$\quad \theta_{(r=0)} = 0$ and $\left(\frac{d \theta}{d r}\right)_{(r=1)} = 0\quad$are not compatible to a non-nul solution $\theta(r)$.