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:
Answer & explanation
Correct answer: option 2
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.