If $y=e^{4 x}+2 e^{-x}$ satisfies the relation $\frac{d^3 y}{d x^3}+A \frac{d y}{d x}+B y=0$ then value of A and B respectively are:

–13, –12

On differentiating $y=e^{4 x}+2 e^{-x}$, w.r.t x we get

$\frac{d y}{d x}=4 e^{4 x}-2 e^{-x} \Rightarrow \frac{d^2 y}{d x^2}=16 e^{4 x}+2 e^{-x} \Rightarrow \frac{d^3 y}{d x^3}=64 e^{4 x}-2 e^{-x}$

Putting these values in $\frac{d^3 y}{d x^3}+A \frac{d y}{d x}+B y=0$, we get

$(64+4 A+B) e^{4 x}+(-2-2 A+2 B) e^{-x}=0$

On solving we get A = –13 and B = –12

Hence (2) is the correct answer.