Answer
$y=(\theta^{2})(\sec \theta)+(\dfrac{18}{\pi^2}-2)\theta^{2}$
Work Step by Step
Here, we have $y'-\dfrac{2y}{\theta}=\theta^2 (\sec \theta)(\tan \theta)$
The integrating factor is: $I=e^{\int -2 \ln |\theta| d\theta}=\theta^{-2}$
Now, $\theta^{-2}[y'-\dfrac{2y}{\theta}]=\theta^{-2}[\theta^2 (\sec \theta)(\tan \theta)]$
$\implies y=(\theta)^{2}(\sec \theta)+c(\theta)^{2}$
Next, apply the initial conditions in the above equation, we have
$ c=(\dfrac{18}{\pi^2})-2$
Hence, $y=(\theta^{2})(\sec \theta)+(\dfrac{18}{\pi^2}-2)\theta^{2}$
