Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 10 - Introduction to Differential Equations - 10.1 Solving Differential Equations - Exercises - Page 504: 12


(a) $y^{3} d y =9 x^{2} d x$ (b) $\frac{y^{4}}{4} =3 x^{3}+C$ (c) $ y= (12 x^{3}+C)^{1/4}$ (d) $y= (12 x^{3}+4)^{1/4}$

Work Step by Step

Given $$y^{3} y^{\prime}-9 x^{2}=0$$ (a) Then \begin{aligned} y^{3} \cdot \frac{d y}{d x} &=9 x^{2} \\ y^{3} d y &=9 x^{2} d x \end{aligned} (b) Integrate both sides \begin{aligned} \int y^{3} d y &=\int 9 x^{2} d x \\ \frac{y^{4}}{4} &=9 \cdot \frac{x^{3}}{3}+C \\ \frac{y^{4}}{4} &=3 x^{3}+C \end{aligned} (c) Then $$ y= (12 x^{3}+C)^{1/4}$$ (d) At $x= 1$, $y=2$, then $C=4$, hence $$ y= (12 x^{3}+4)^{1/4}$$
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