Chapter 8 - Linear Differential Equations of Order n - 8.1 General Theory for Linear Differential Equations - Problems - Page 504: 37

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Given: $x^3y'''+3x^2y''-6xy'=0$ Substitute: $x^3y'''(x)+3x^2y''(x)-6xy'(x)\\ =x^3r(r-1)(r-2)x^{r-3}+3x^2r(r-1)x^{r-2}-6xrx^{r-1}\\ =x^r(r^3-3r^2+2r+3r^2-3r-6r)\\ =r^3-7r\\ =0$ Since $e^{rx} \ne0 \rightarrow r^3-7r=0\\ \rightarrow r(r^2-7)=0\\ \rightarrow r=0,r=\pm \sqrt 7$ The solutions to the given problem are: $y_1(x)=C_1x\\ y_2(x)=C_2x^{\sqrt 7}\\ y_3(x)=C_3x^{-\sqrt 7}$