Chapter 8 - Linear Differential Equations of Order n - 8.1 General Theory for Linear Differential Equations - Problems - Page 504: 35

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Given: $2x^2y''+5xy'+y=0$ Substitute: $2x^2y''(x)+5xy'(x)+y(x)\\ =2r^2x^2(x^r)^2+5x^r+x^r\\ =x^r(2r^2-2r+5r+1)\\ =2r^2+3r+1\\ =0$ Since $e^{rx} \ne0 \rightarrow r^2+3r+1=0\\ \rightarrow (2r+1)(r+1)=0\\ \rightarrow r=-\frac{1}{2}, r=-1$ The solutions to the given problem are: $y_1(x)=C_1x^{-\frac{1}{2}}\\ y_2(x)=C_2x^{-1}$