If y=e−xcos⁡x and yn+kny=0, where yn=dnydxn and kn are , constant ∀n∈N then

y=e−xcos⁡xy1=−e−xcos⁡x−e−xsin⁡x=−2e−xcos⁡x−π4y2=(−2)2e−xcos⁡x−π2y3=(−2)3e−xcos⁡x−3π4y4=(−2)4e−xcos⁡(x−π)=−4e−xcos⁡x or y4+4y=0 or k4=4 Differentiating it again four times, we get y8+4y4=0 or y8−16y=0 or k8=−16y12+4y8=0 or y12+64y=0 or k12=64; Similarly ,k12=−256

If y=e−xcos⁡x and yn+kny=0, where yn=dnydxn and kn are , constant ∀n∈N then

y=e−xcos⁡xy1=−e−xcos⁡x−e−xsin⁡x=−2e−xcos⁡x−π4y2=(−2)2e−xcos⁡x−π2y3=(−2)3e−xcos⁡x−3π4y4=(−2)4e−xcos⁡(x−π)=−4e−xcos⁡x or y4+4y=0 or k4=4 Differentiating it again four times, we get y8+4y4=0 or y8−16y=0 or k8=−16y12+4y8=0 or y12+64y=0 or k12=64; Similarly ,k12=−256

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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 , constant $\forall n \in N$ then

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$k_{4} = 4$

$k_{8} = - 16$

$k_{12} = 20$

$k_{16} = - 24$

answer is A, B.

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Detailed Solution

$\begin{array}{l} y = e^{- x} \cos x \\ y_{1} = - e^{- x} \cos x - e^{- x} \sin x = - \sqrt{2} e^{- x} \cos (x - \frac{π}{4}) \\ y_{2} = (- \sqrt{2})^{2} e^{- x} \cos (x - \frac{π}{2}) \\ y_{3} = (- \sqrt{2})^{3} e^{- x} \cos (x - \frac{3 π}{4}) \\ y_{4} = (- \sqrt{2})^{4} e^{- x} \cos (x - π) = - 4 e^{- x} \cos x \\ or y_{4} + 4 y = 0 or k_{4} = 4 \end{array}$