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

A, B

Explanation

`y=e^(-x)cosx`
`therefore" "y_(1)=-e^(-x)cosx-e^(-x)sinx`
`=-sqrt2e^(-x)cos(x-(pi)/(4))`
Similarly, we get
`y_(2)=(-sqrt2)^(2)e^(-x)cos(x-(pi)/(2))`
`y_(3)=(-sqrt2)^(3)e^(-x)cos(x-(3pi)/(4))`
`y_(4)=(-sqrt2)^(4)e^(-x)cos(x-pi)=-4e^(-x)cosx`
`rArr" "y_(4)+4y=0`
`rArr" "k_(4)=4`
Differentiating it again four times, we get
`y_(8)+4y_(4)=0`
`rArr" "y_(8)-16y=0 rArr k_(8)=-16`
`y_(12)+4y_(8)=0 rArr y_(12)+64y=0 rArr k_(12)=64`
Similarly `k_(16)=-256`