If  $y=e^{4x}+2e^{-x}$ satisfies the differential equation $y_3+A{y}_1+By=0$ then 

$$\begin{gathered} y=e^{4x}+2e^{-x} \\ diffw.r.tx \\ \Rightarrow \frac{dy}{dx}=4e^{4x}-2e^{-x} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=16e^{4x}+2e^{-x} \\ \Rightarrow \frac{d^{3}y}{d{x}^3}=64e^{4x}-2e^{-x} \\ puttingthesevaluesin\frac{d^{3}y}{d{x}^3}+A\frac{dy}{dx}+By=0 \\ thenweget64+4A+B=0,-2-2A+2B=0 \\ By solving these,wegetA=-13,B=-12 \end{gathered}$$