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: 17

Answer

Answer given below.

Work Step by Step

Take the derivatives of the function. $y(x)=c_1x^{\frac{1}{2}}+3x^2$ $y'(x)=\frac{1}{2}c_1x^{\frac{-1}{2}}+6x$ $y''(x)=\frac{-1}{4}c_1x^{\frac{-3}{2}}+6$ Substituting these functions into the differential equation yields $\implies 2x^2y''-xy'+y=9x^2$ $\implies 2x^2(\frac{-1}{4}c_1x^{\frac{-3}{2}}+6)-x(\frac{1}{2}c_1x^{\frac{-1}{2}}+6x)+c_1x^{\frac{1}{2}}+3x^2=9x^2$ $\implies \frac{-1}{2}c_1x^{\frac{1}{2}}+12x^2-\frac{1}{2}c_1x^{\frac{1}{2}}-6x^2+c_1x^{\frac{1}{2}}+3x^2-9x^2=0$ $\implies 0=0$ This means that the solution is valid for all real values. The interval linked to this solution is $(−∞,∞)$.
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