Question

If the function $y=e^{4 x}+2 e^{-x}$ is a solution of the differential equation $\frac{\frac{d^3 y}{d x^3}-13 \frac{d y}{d x}}{y}=k$, then the value of $\frac{k}{3}$ is

• A. -4
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (D)

$y=e^{4 x}+2 e^{-x}$
$\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} $
$\frac{d^3 y}{d x^3}=64 e^{4 x}-2 e^{-x}$
$\therefore \frac{\frac{d^3 y}{d x^3}-13 \frac{d y}{d x}}{y}=\frac{\left(64 e^{4 x}-2 e^{-x}\right)-13\left(4 e^{4 x}-2 e^{-x}\right)}{e^{4 x}+2 e^{-x}}=\frac{12 e^{4 x}+24 e^{-x}}{e^{4 x}+2 e^{-x}}=12 $
$\therefore \frac{k}{3}=4 $