Answer to Question #224868 in Chemical Engineering for Lokika

Question

Solve (x²D² + 4xD + 2)y= (logx/x)^2,x> 0, where D =d/dx.

Answer 1

The auxiliary equation is

$m^2+4m+2=0\\ m_{1,\:2}=\frac{-4\pm \sqrt{4^2-4\cdot \:1\cdot \:2}}{2\cdot \:1}\\ m=-2+\sqrt{2},\:m=-2-\sqrt{2}$

Thus the complementary solution is

$y_c= (c_1+xc_2)e^{-x}\\ Let \space y_1= e^{-x} ; y_2= xe^{-x}\\ w= \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2' \end{vmatrix}= \begin{vmatrix} e^{-x} & xe^{-x}\\ -e^{-x} & -xe^{-x}+e^{-x} \end{vmatrix}= e^{-2x}\\$

By the method of variation of parameter particular solution is

$y_p=-y_1 \int \frac{y_2 f(x)}{w}dx+y_2 \int \frac{y_1 f(x)}{w}dx\\ y_p=x^2e^{-x}[\frac{log x }{2}+ \frac{1}{4}+log x -1]\\ y_p= \frac{x^2e^{-x}}{4}[2log x -3]\\$

Hence the general solution is

$y(x)= (c_1+xc_2)e^{-x}+\frac{x^2e^{-x}}{4}[2log x -3]$