Chapter 8 - Linear Differential Equations of Order n - 8.1 General Theory for Linear Differential Equations - Problems - Page 504: 33

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Given: $y^{iv}-13y''+36y=0$ Substitute: $y^{iv}-13y''(x)+36y(x)\\ =r^4e^{rx}-13r^2e^{rx}+36re^{rx}\\ =e^{rx}(r^4-13r^2+36)\\ =0$ Since $e^{rx} \ne0 \rightarrow r^4-13r^2+36=0\\ \rightarrow (r^2-9)(r^2-4)=0\\ \rightarrow r=\pm 3, r= \pm 2$ The solutions to the given problem are: $y_1(x)=C_1e^{3x}\\ y_2(x)=C_2e^{-3x} \\ y_3(x)=C_3e^{2x} \\ y_4(x)=C_4e^{-2x}$