Question

If $y=e^{-x}\cos x$ and $y_n+k_{n}y=0$, where $y_n=\frac{d^{n}y}{d{x}^n}$ and $k_n$ are constants $\forall nϵN$, then

• A. $k_4=4$
• B. $k_8=-16$
• C. $k_{12}=20$
• D. $k_{16}=-24$

Answer 1

Answer: (A), (B)

The correct options are
A $k_4=4$
B $k_8=-16$

$y=e^{-x}\cos x$
$y_1=-e^{-x}\cos x-e^{-x}\sin x=-\sqrt{2}{e}^{-x}\cos (x-\frac{\pi }{4})$
$y_2={(-\sqrt{2})}^2{e}^{-x}\cos (x-\frac{\pi }{2})$
$y_3={(-\sqrt{2})}^3{e}^{-x}\cos (x-\frac{3\pi }{4})$
$y_4={(-\sqrt{2})}^4{e}^{-x}\cos (x-\pi )=-4e^{-x}\cos x$
or $y_4+4y=0$ or $k_4=4$
Differentiating it again four times, we get
$y_8+4y_4=0$
or $y_8-16y=0$
or $k_8=-16$
$y_{12}+4y_8=0$
or $y_{12}+64y=0$
or $k_{12}=64$
Similarly, $k_{16}=-256$