Calculus 10th Edition

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

Chapter 6 - Differential Equations - 6.1 Exercises - Page 403: 24


$y=x^2(2+e^x)$ is a solution to the given differential equation.

Work Step by Step

We will find $y'$ and put it into the differential equation along with $y$. $$y'= (x^2(2+e^x))'= (x^2)'(2+e^x)+x^2(2+e^x)'=2x(2+e^x)+x^2((2)'+(e^x)') = 4x+2xe^x+x^2e^x$$ Putting this into the differential equation we get The Left side: $$x(4x+2xe^x+x^2e^x)-2x^2(2+e^x) = 4x^2+2x^2e^x+x^3e^x-4x^2-2x^2e^x = x^3e^x.$$ The Right side is just $$x^3e^x.$$ We see that the Left side is equal to the Right side so $y=x^2(2+e^x)$ IS a solution to this differential equation.
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