Answer
$y=-\dfrac{\cos \theta}{\theta}+\dfrac{\pi}{2\theta}$
Work Step by Step
The standard form of the given equation is:
$y'+\dfrac{1}{\theta}y=\dfrac{\sin \theta}{\theta}$ ....(1)
The integrating factor is: $e^{\int \frac{1}{\theta} d\theta}=\theta$
In order to determine the general solution, multiply equation (1) with the integrating factor and integrate both sides.
$\int [y\theta]' dt=\int (\dfrac{\sin \theta}{\theta})(\theta) d\theta$
or, $y\theta=-\cos \theta+c$
Applying the initial conditions in the above equation, we get
$(1)(\pi/2)=-\cos(\pi/2)+c \implies c=\dfrac{\pi}{2}$
Then $y\theta=-\cos \theta+\dfrac{\pi}{2}$
Thus, we have $y=-\dfrac{\cos \theta}{\theta}+\dfrac{\pi}{2\theta}$
