## University Calculus: Early Transcendentals (3rd Edition)

a) $y=\tan \theta +C$ b) $y=\dfrac{2}{3} \sqrt{\theta^{3}}+C$ c) $y=\dfrac{2}{3} \sqrt{\theta^{3}}-\tan \theta +C$
a) When $\sec^2 \theta=(tan \theta)'$ Thus, $y'=\sec^2 \theta\implies y=\tan \theta +C$ b) When $\sqrt {\theta}=(\dfrac{2}{3} \sqrt{\theta^{3}})'$ Thus, $y'=\sqrt {\theta}\implies y=\dfrac{2}{3} \sqrt{\theta^{3}}+C$ c) The general solution will be the difference of part a and part b, that is, $y'=\dfrac{2}{3} \sqrt{\theta^{3}}-\tan \theta +C$