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# Answer to Question #224868 in Chemical Engineering for Lokika

Question #224868

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

1
2021-08-16T02:19:50-0400

The auxiliary equation is

"m^2+4m+2=0\\\\\nm_{1,\\:2}=\\frac{-4\\pm \\sqrt{4^2-4\\cdot \\:1\\cdot \\:2}}{2\\cdot \\:1}\\\\\nm=-2+\\sqrt{2},\\:m=-2-\\sqrt{2}"

Thus the complementary solution is

"y_c= (c_1+xc_2)e^{-x}\\\\\nLet \\space y_1= e^{-x} ; y_2= xe^{-x}\\\\\nw= \\begin{vmatrix}\n y_1 & y_2 \\\\\n y_1' & y_2'\n\\end{vmatrix}= \\begin{vmatrix}\n e^{-x} & xe^{-x}\\\\\n -e^{-x} & -xe^{-x}+e^{-x}\n\\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\\\\\ny_p=x^2e^{-x}[\\frac{log x }{2}+ \\frac{1}{4}+log x -1]\\\\\ny_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]"

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