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If $y=e^{4 x}+2 e^{-x}$ satisfies the equation $\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+A \frac{\mathrm{d} y}{\mathrm{~d} x}+\mathrm{By}=o$ then values of and $B$ are respectively
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-3,-4
$\begin{aligned} & y=e^{4 x}+2 e^{-x} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=4 \cdot e^{4 x}-2 \cdot e^{-x} \\ & \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=16 e^{4 x}+2 e^{-x} \\ & \Rightarrow \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-3 \frac{\mathrm{d} y}{\mathrm{~d} x}-4 y=0 \\ & \Rightarrow A=-3 \text { and } B=-4\end{aligned}$
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