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 505: 33



Work Step by Step

By separation of variables, we have $$ye^{y^2}dy=xdx$$ then by integration, we get $$\frac{1}{2}e^{y^2}=\frac{1}{2}x^2+c\Longrightarrow y^2=\ln(x^2+A).$$ Now, since $y(0)=-2$, then $4=\ln(0+A)\Longrightarrow A= e^{4}$. So the general solution is given by $$y^2=\ln(x^2+e^4).$$ We take the square root to get: $$y=-\sqrt{\ln(x^2+e^4)}.$$ Note that we use the negative value because we know that $y(0)$ is negative.
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