Differential Equations and Linear Algebra (4th Edition)

Published by Pearson
ISBN 10: 0-32196-467-5
ISBN 13: 978-0-32196-467-0

Chapter 1 - First-Order Differential Equations - 1.4 Separable Differential Equations - Problems - Page 43: 13

Answer

\[y(x)=a(1+\sqrt{1-x^2})\]

Work Step by Step

$(1-x^2)y'+xy=ax$ _____(1) $(1-x^2)y'=x(a-y)$ $(1-x^2)\frac{dy}{dx}=x(a-y)$ Separating variables, $\frac{x}{1-x^2}dx=\frac{dy}{a-y}$ Integrating, $C+\int\frac{x}{1-x^2}dx=\int\frac{dy}{a-y}$ $C$ is constant of integration $C-\frac{1}{2}\int\frac{-2x}{1-x^2}dx=-\ln|a-y|$ $C-\frac{1}{2}\ln|1-x^2|=\ln\left|\frac{1}{a-y}\right|$ $\ln\left|\frac{\sqrt{1-x^2}}{a-y}\right|=C$ Using initial condition $y(0)=2a$ $\ln\left|\frac{1}{a-2a}\right|=C$ $\ln\left|\frac{-1}{a}\right|=C$ ___(3) Using (2) and (3) $\frac{\sqrt{1-x^2}}{a-y}=\frac{-1}{a}$ $\Rightarrow a\sqrt{1-x^2}=y-a$ $\Rightarrow a(1+\sqrt{1-x^2})=y(x)$ Hence solution fo given initial value problem is $y(x)=a(1+\sqrt{1-x^2})$
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