Answer
$a) y=-\dfrac{\cos 2t}{2} +C; b) y=2 \sin \dfrac{t}{2} +C; c) y=-\dfrac{\cos 2t}{2} +2 \sin \dfrac{t}{2}+C$
Work Step by Step
a) When $(\sin 2t)=(-\dfrac{\cos 2t}{2})'$
This implies, $y'= \sin 2t$ and $y=-\dfrac{\cos 2t}{2} +C$
b) When $(\cos \dfrac{t}{2})=( 2 \sin \dfrac{t}{2})'$
This implies, $y'= (\cos \dfrac{t}{2})$ and $y=(2 \sin \dfrac{t}{2}) +C$
c) Here, $y'=\sin 2t+\cos \dfrac{t}{2}$
This implies, $y'=(-\dfrac{\cos 2t}{2} +2 \sin \dfrac{t}{2})+C$
