Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 6 - Differential Equations - 6.3 Exercises: 25

Answer

$$y=\sqrt{\frac{x^2}{4}+4}$$

Work Step by Step

$y'=\frac{x}{4y}$ $4yy'=x$ $\int 4ydy=\int xdx$ $2y^2=\frac{x^2}{2}+C$ $y=\sqrt{\frac{x^2}{4}+C}$ Now we need this equation to pass through $(0,2)$ so we apply it as an initial condition: $2=\sqrt{0+C}$ $C=4$ So our equation becomes $$y=\sqrt{\frac{x^2}{4}+4}$$
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