Chapter 8 - Linear Differential Equations of Order n - 8.1 General Theory for Linear Differential Equations - Problems - Page 504: 34

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Given: $x^2y''+3xy'-8y=0$ Substitute: $x^2y''(x)+3xy'(x)-8y(x)\\ =r^2x^2(x^r)^2+3x^r-8x^r\\ =x^r(r^2-r+3r-8)\\ =r^2+2r-8\\ =0$ Since $e^{rx} \ne0 \rightarrow r^2+2r-8=0\\ \rightarrow (r+4)(r-2)=0\\ \rightarrow r=-4, r= \pm 2$ The solutions to the given problem are: $y_1(x)=C_1x^{-4}\\ y_2(x)=C_2x^2$