Chapter 8 - Linear Differential Equations of Order n - 8.1 General Theory for Linear Differential Equations - Problems - Page 504: 36

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Given: $x^3y'''+x^2y''-2xy'+2y=0$ Substitute: $x^3y'''(x)+x^2y''(x)-2xy'(x)+2y(x)\\ =x^3r(r-1)(r-2)x^{r-3}+x^2r(r-1)x^{r-2}-2xrx^{r-1}+2x^r\\ =x^r(r^3-3r^2+2r+r^2-r-2r+2)\\ =r^3-2r^2-r+2\\ =0$ Since $e^{rx} \ne0 \rightarrow r^3-2r^2-r+2=0\\ \rightarrow (r-1)(r^2-r-2)=0\\ \rightarrow r=1, r=-1,r=2$ The solutions to the given problem are: $y_1(x)=C_1x\\ y_2(x)=C_2x^2\\ y_3(x)=C_3x^{-1}$