Answer
$y=-\dfrac{\cos \theta}{\theta}+\dfrac{\pi}{2\theta}$
Work Step by Step
Here, we have $y'+\dfrac{1}{\theta}y=\dfrac{\sin \theta}{\theta}$
The integrating factor is: $I=e^{\int \frac{1}{\theta} d\theta}=\theta$
Now, $\theta[y'+\dfrac{1}{\theta}y]=\theta[\dfrac{\sin \theta}{\theta}]$
This gives: $\int [y\theta]' dt=\int (\dfrac{\sin \theta}{\theta})(\theta) d\theta$
$\implies (y)(\theta)=-(\cos \theta)+c$
Next, apply the initial conditions in the above equation, we have
$(1)(\dfrac{\pi}{2})=-\cos(\dfrac{\pi}{2})+c$
or, $c=\dfrac{\pi}{2}$
Thus, $(y)(\theta)=-(\cos \theta)+\dfrac{\pi}{2}$
Hence, $y=-\dfrac{\cos \theta}{\theta}+\dfrac{\pi}{2\theta}$
