Chapter 4 - Integration - 4.2 The Indefinite Integral - Exercises Set 4.2 - Page 279: 40

$$\eqalign{ & \left. {\text{a}} \right)y = - \frac{1}{{16{x^2}}} + \frac{1}{{16}} \cr & \left. {\text{b}} \right)y = \tan t + \cos t - \frac{{\sqrt 2 }}{2} \cr & \left. {\text{c}} \right)y = \frac{2}{9}{x^{9/2}} \cr} $$

$$\eqalign{ & \left. {\text{a}} \right){\text{ }}\frac{{dy}}{{dx}} = \frac{1}{{{{\left( {2x} \right)}^3}}},{\text{ }}y\left( 1 \right) = 0 \cr & \frac{{dy}}{{dx}} = \frac{1}{{8{x^3}}} \cr & \frac{{dy}}{{dx}} = \frac{1}{8}{x^{ - 3}} \cr & dy = \frac{1}{8}{x^{ - 3}}dx \cr & y = \frac{1}{8}\int {{x^{ - 3}}} dx \cr & y = \frac{1}{8}\left( {\frac{{{x^{ - 2}}}}{{ - 2}}} \right) + C \cr & y = - \frac{1}{{16{x^2}}} + C \cr & {\text{Using the initial condition}} \cr & 0 = - \frac{1}{{16{{\left( 1 \right)}^2}}} + C \cr & C = \frac{1}{{16}} \cr & {\text{then}} \cr & y = - \frac{1}{{16{x^2}}} + \frac{1}{{16}} \cr & \cr & \left. {\text{b}} \right){\text{ }}\frac{{dy}}{{dt}} = {\sec ^2}t - \sin t,{\text{ }}y\left( {\frac{\pi }{4}} \right) = 1 \cr & dy = \left( {{{\sec }^2}t - \sin t} \right)dt \cr & y = \int {\left( {{{\sec }^2}t - \sin t} \right)} dt \cr & y = \tan t + \cos t + C \cr & {\text{Using the initial condition}} \cr & 1 = \tan \left( {\frac{\pi }{4}} \right) + \cos \left( {\frac{\pi }{4}} \right) + C \cr & 1 = 1 + \frac{{\sqrt 2 }}{2} + C \cr & C = - \frac{{\sqrt 2 }}{2} \cr & {\text{then}} \cr & y = \tan t + \cos t - \frac{{\sqrt 2 }}{2} \cr & \cr & \left. {\text{c}} \right){\text{ }}\frac{{dy}}{{dx}} = {x^2}\sqrt {{x^3}} ,{\text{ }}y\left( 0 \right) = 0 \cr & \frac{{dy}}{{dx}} = {x^{7/2}} \cr & dy = {x^{7/2}}dx \cr & y = \int {{x^{7/2}}} dx \cr & y = \frac{2}{9}{x^{9/2}} + C \cr & {\text{Using the initial condition}} \cr & 0 = \frac{2}{9}{\left( 0 \right)^{9/2}} + C \cr & C = 0 \cr & {\text{then}} \cr & y = \frac{2}{9}{x^{9/2}} \cr} $$