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 - 1.2 Basic Ideas and Terminology - Problems - Page 22: 49

Answer

See answer below

Work Step by Step

Take the derivatives of the function. $$y(x)=c_1e^x+c_2e^{-x}+(2x^2+2x+1)$$ $$y'(x)=c_1e^x-c_2e^{-x}+(2x^2+2x+1)+c_2e^{-x}(4x+2)$$ $$y''(x)=c_1e^x+c_2e^{-x}+(2x^2+2x+1)-2c_2e^{-x}(4x+2)+4c_2e^{-x}$$ Substitute these functions into the differential equation to get $$xy''-2y'+(2-x)y=0$$ $$x[c_1e^x+c_2e^{-x}+(2x^2+2x+1)-2c_2e^{-x}(4x+2)+4c_2e^{-x}]-2[c_1e^x-c_2e^{-x}+(2x^2+2x+1)+c_2e^{-x}(4x+2)]+(2-x)[c_1e^x+c_2e^{-x}+(2x^2+2x+1)]=0$$ $$0=0$$ This equation is always true, so the equation is a solution to the differential equation.
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