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 - 1.6 First-Order Linear Differential Equations - Problems - Page 59: 4

Answer

$y=x^2-1+Ce^{-x^{2}}$

Work Step by Step

Given that: $y'+2xy=2x^3$ Use an integration factor of: $$\mu (x)=e^{\int P(x)dx}$$ where $P(x)=2x$ $$\mu (x)=e^{\int 2xdx}=e^{x^2}$$ Multiply the entire equation by this factor. $$e^{x^2}(y'+2xy=2x^3)$$$$e^{x^2}y'+2e^{x^2}xy=2e^{x^2}x^3$$ Integrate each side. $$\int e^{x^2}y'+2e^{x^2}xy=\int 2e^{x^2}x^3 dx$$ To evluate the intergral $\int 2e^{x^2}x^3 dx$ let $t=x^2 \rightarrow dt=2xdx$ The intergral become $\int te^tdt=te^t-\int e^tdt=te^t-e^t=e^t(t-1) \rightarrow \int 2e^{x^2}x^3 dx=e^{x^2}(x^2-1)$ Thus; $$ye^{x^2}=e^{x^2}(x^2-1)+C$$ $$y=x^2-1+Ce^{-x^{2}}$$
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