Answer
$y=(\theta^{2})(\sec \theta)+(\dfrac{18}{\pi^2}-2)\theta^{2}$
Work Step by Step
The standard form of the given equation is:
$y'-\dfrac{2y}{\theta}=\theta^2 (\sec \theta)(\tan \theta)$ ....(1)
The integrating factor is: $e^{\int -2 \ln |\theta| d\theta}=\theta^{-2}$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$y=(\theta^{2})(\sec \theta)+c(\theta^{2})$ ...(2)
Applying the initial conditions in the above equation, we get
$ c=\dfrac{18}{\pi^2}-2$
Thus, equation (2) becomes: $y=(\theta^{2})(\sec \theta)+(\dfrac{18}{\pi^2}-2)\theta^{2}$
