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: 2

Answer

$y(x)=c_1e^x+c_2e^{-x}-x^3-5x$

Work Step by Step

The auxiliary equation is: $r^2-1=0$ We factor: $(r+1)(r-1)=0$ Thus $r=\pm 1$ Therefore, the complementary equation is: $y_c(x)=c_1*e^x+c_2*e^{-x}$ We attempt: $y_p(x)=Ax^3+Bx^2+Cx+D$ Thus: $y_p'(x)=3Ax^2+2Bx$ $y_p''(x)=6Ax+2B$ We plug into the differential equation: $(6Ax+2B)-(Ax^3+Bx^2+Cx+D)=x^3-x$ If we examine the coefficients, we get: $A=-1, B=0, C=-5, D=0$ Therefore, the general solution is: $y(x)=c_1e^x+c_2e^{-x}-x^3-5x$
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