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 21: 18

Answer

Given below.

Work Step by Step

Take the derivatives of the function. $y(x)=c_1x^2+c_2x^3-x^2sinx$ $y'(x)=2xc_1+3x^2c_2-2xsinx - x^2cosx$ $y''(x)=2c_1+6xc_2-2sinx-4xcosx+x^2sinx$ Substituting these functions into the differential equation yields $x^2y'' -4xy' +6y= x^4sinx$ $x^2(2c_1+6xc_2-2sinx-4xcosx+x^2sinx)-4x(2xc_1+3x^2c_2-2xsinx - x^2cosx)+6(c_1x^2+c_2x^3-x^2sinx)=x^4sinx$ $2c_1x^2+6c_2x^3-2x^2sinx-4x^3cosx+x^4sinx-8c_1x^2-12c_2x^3+8x^2sinx+4x^3cosx+6c_1x^2+6c_2x^3-6x^2sinx-x^4sinx=0$ $0=0$ This means that the solution is valid for all real values. The interval linked to this solution is $(−∞,∞)$.
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