Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 17 - Second-Order Differential Equations - 17.2 Exercises - Page 1179: 17

Answer

$y_p(x)=xe^{-x}[(A x^2+Bx+C)\cos 3x+(D x^2+Ex+F)\sin 3x]$

Work Step by Step

Consider $G(x)=e^{ax} A(x) \sin mx $ or $G(x)=e^{ax} A(x) \cos mx $ The trial solution for the method of undetermined coefficients can be calculated as: $y_p(x)=e^{ax} B(x) \sin mx +e^{ax} C(x) \cos mx$ when the sum of the coefficients of a differential equation is zero. Then, we have $y_p(x)=x e^{ax} B(x)$ Given: $y''+2y'+10y=x^2e^{-x}\cos 3x$ Here, we have $m=k=1$ Thus, the trial solution for the method of undetermined coefficients is: $y_p(x)=xe^{-x}(A x^2+Bx+C)\cos 3x+xe^{-x}(D x^2+Ex+F)\sin 3x$ Hence, $y_p(x)=xe^{-x}[(A x^2+Bx+C)\cos 3x+(D x^2+Ex+F)\sin 3x]$
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