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: 47

Answer

See answer below

Work Step by Step

Take the derivatives of the function. $$y(x)=c_1x^2+c_2x^{2}\ln (x)+\frac{1}{6}x^2(\ln(x))^3$$ $$y'(x)=\frac{1}{6}x(12c_2\ln(x)+12c_1+6c_2+2 \ln^3(x)+3\ln^2(x))$$ $$y''(x)=2c_2\ln (x)+3c_2+2c_1+\frac{1}{3}\ln^3(x)+\frac{3}{2}\ln^2(x)+\ln (x)$$ Substitute these functions into the differential equation to get $$x^2y''+4xy=0$$ $$x^2[2c_2\ln (x)+3c_2+2c_1+\frac{1}{3}\ln^3(x)+\frac{3}{2}\ln^2(x)+\ln (x)]-3x[\frac{1}{6}x(12c_2\ln(x)+12c_1+6c_2+2 \ln^3(x)+3\ln^2(x))]+4[c_1x^2+c_2x^{2}\ln (x)+\frac{1}{6}x^2(\ln(x))^3]=x^2\ln(x)$$ $$\frac{\ln(x)x^2}{6}[2\ln^2(x)+9\ln(x)+6]-\frac{3x^2}{6}[2\ln^3(x)+3\ln^2(x)]+\frac{4x^2}{6}=x^2\ln(x)$$ $$\ln(x)x^2=x^2\ln(x)$$ $$0=0$$ This equation is always true, so the equation is a solution to the differential equation.
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