Answer to Question #319378 in Differential Equations for Sharlene

Question

Obtain the solution of the differential equation

(x^2)y" - 2xy' + 2y = (log (base e) x) ^2, (x>0),

using the method of undetermined coefficients.

Answer 1

$x^2y''-2xy'+2y=\ln ^2x\\Homogeneous\,\,equation:\\x^2y''-2xy'+2y=0\\y=zx\\y'=z+z'x\\y''=2z'+z''x\\2z'x^2+z''x^3-2zx-2z'x^2+2zx=0\\z''x^3=0\\z''=0\\z=C_1+C_2x\\y=C_1x+C_2x^2\\The\,\,particular\,\,solution\,\,of\,\,the\,\,inhomogeneous\,\,equation\,\,\\y=A\ln x+B\ln ^2x+C\\y'=\frac{A}{x}+2B\frac{\ln x}{x}\\y''=-\frac{A}{x^2}+\frac{2B}{x^2}-\frac{2B\ln x}{x^2}\\-A+2B-2B\ln x-2A-2B\ln x+2A\ln x+2B\ln ^2x+2C=\ln ^2x\\\left\{ \begin{array}{c} -3A+2B+2C=0\\ -4B+2A=0\\ 2B=1\\\end{array} \right. \Rightarrow \left\{ \begin{array}{c} A=\frac{3}{2}\\ B=\frac{1}{2}\\ C=\frac{7}{4}\\\end{array} \right. \\y=C_1x+C_2x^2+\frac{3}{2}\ln x+\frac{1}{2}\ln ^2x+\frac{7}{4}$