If `y=e^(-x)cosxa n dy_n+k_n y=0,w h e r ey_n=(d^(n y))/(dx^n)a n dk_n` are constants `AAn in N ,` then `k_4=4` (b) `k_8=-16` `k_(12)=20` (d) `k_(16)=-24`

Options

A. `k_(4)=4`
B. `k_(8)=-16`
C. `k_(12)=20`
D. `k_(16)=-24`

Answer

B

Explanation

`y=e^(-x)cos x`
`y_(1)=-e^(-x)cos x - e^(-x)sin x=-sqrt(2) e^(-x)cos (x-(pi)/(4))`
`y_(2)=(-sqrt(2))e^(-x)cos (x-(pi)/(2))`
`y_(3)=(-sqrt(2))^(3)e^(-x)cos (x-(3pi)/(4))`
`y_(4)=(-sqrt(2))^(4)e^(-x)cos (x-pi)=-4 e^(-x)cos x`
`"or "y_(4)+4y=0or k_(4)=4`
Differentiating it again four times, we get
`y_(8)+4y_(4)=0`
`"or "y_(8)-16y=0`
`"or "k_(8)=-16`
`"Further "y_(12)+4y_(8)=0`
`"or "y_(12)+64y=0`
`"or "k_(12)=64`
`"Similarly, "k_(16)=-256`