Chapter 4 - Integration - 4.2 The Indefinite Integral - Exercises Set 4.2 - Page 279: 39

$$\eqalign{ & \left. {\text{a}} \right)y = \frac{3}{4}{x^{4/3}} + \frac{5}{4} \cr & \left. {\text{b}} \right)y = - \cos t + t + 1 - \frac{\pi }{3} \cr & \left. {\text{c}} \right)y = \frac{2}{3}{x^{3/2}} + 2{x^{1/2}} - \frac{8}{3} \cr} $$

$$\eqalign{ & \left. {\text{a}} \right){\text{ }}\frac{{dy}}{{dx}} = \root 3 \of x ,{\text{ }}y\left( 1 \right) = 2 \cr & dy = {x^{1/3}}dx \cr & y = \int {{x^{1/3}}} dx \cr & y = \frac{3}{4}{x^{4/3}} + C \cr & {\text{Using the initial condition}} \cr & 2 = \frac{3}{4}{\left( 1 \right)^{4/3}} + C \cr & C = \frac{5}{4} \cr & {\text{then}} \cr & y = \frac{3}{4}{x^{4/3}} + \frac{5}{4} \cr & \cr & \left. {\text{b}} \right){\text{ }}\frac{{dy}}{{dt}} = \sin t + 1,{\text{ }}y\left( {\frac{\pi }{3}} \right) = \frac{1}{2} \cr & dy = \left( {\sin t + 1} \right)dt \cr & y = \int {\left( {\sin t + 1} \right)} dt \cr & y = - \cos t + t + C \cr & {\text{Using the initial condition}} \cr & \frac{1}{2} = - \cos \left( {\frac{\pi }{3}} \right) + \frac{\pi }{3} + C \cr & \frac{1}{2} = - \frac{1}{2} + \frac{\pi }{3} + C \cr & C = 1 - \frac{\pi }{3} \cr & {\text{then}} \cr & y = - \cos t + t + 1 - \frac{\pi }{3} \cr & \cr & \left. {\text{c}} \right){\text{ }}\frac{{dy}}{{dx}} = \frac{{x + 1}}{{\sqrt x }},{\text{ }}y\left( 1 \right) = 0 \cr & dy = \frac{{x + 1}}{{\sqrt x }}dx \cr & y = \int {\left( {{x^{1/2}} + {x^{ - 1/2}}} \right)} dx \cr & y = \frac{2}{3}{x^{3/2}} + 2{x^{1/2}} + C \cr & {\text{Using the initial condition}} \cr & 0 = \frac{2}{3}{\left( 1 \right)^{3/2}} + 2{\left( 1 \right)^{1/2}} + C \cr & C = - \frac{8}{3} \cr & {\text{then}} \cr & y = \frac{2}{3}{x^{3/2}} + 2{x^{1/2}} - \frac{8}{3} \cr} $$