Answer
a) $y=\tan \theta +C$
b) $y=\dfrac{2}{3} \sqrt{\theta^{3}}+C$
c) $y=\dfrac{2}{3} \sqrt{\theta^{3}}-\tan \theta +C$
Work Step by Step
a) When $\sec^2 \theta=(tan \theta)'$
This implies, $y'=\sec^2 \theta$ and $y=\tan \theta +C$
b) When $\sqrt {\theta}=(\dfrac{2}{3}) \sqrt{\theta^{3}}'$
This implies, $y'=\sqrt {\theta}$ and $y=(\dfrac{2}{3}) \sqrt{\theta^{3}}+C$
c) The general solution will be a difference of part a from part b. Therefore, $y'=(\dfrac{2}{3}) \sqrt{\theta^{3}}-\tan \theta +C$
