If `y=e^(-x) cos x` and `y_n+k_ny=0` where `yn=(d^ny)/(dx^n)` and `k_n` are constant `n in N` then

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`