Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 4 - Integration - Chapter 4 Review Exercises - Page 342: 5

Answer

$$\,\left( {\text{a}} \right)y = 2\sqrt x - \frac{2}{3}{x^{3/2}} - \frac{4}{3},\,\,\,\left( {\text{b}} \right)y = \sin x - \frac{5}{2}{x^2} + 1,\,\,\,\left( {\text{c}} \right)y = \frac{3}{4}{x^{4/3}} + \frac{5}{4}$$

Work Step by Step

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