Question #224868

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


1
Expert's answer
2021-08-16T02:19:50-0400

The auxiliary equation is

m2+4m+2=0m1,2=4±4241221m=2+2,m=22m^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

yc=(c1+xc2)exLet y1=ex;y2=xexw=y1y2y1y2=exxexexxex+ex=e2xy_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

yp=y1y2f(x)wdx+y2y1f(x)wdxyp=x2ex[logx2+14+logx1]yp=x2ex4[2logx3]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)=(c1+xc2)ex+x2ex4[2logx3]y(x)= (c_1+xc_2)e^{-x}+\frac{x^2e^{-x}}{4}[2log x -3]


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