InterviewSolution
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InterviewSolution
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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` thenA. `k_(4)=4`B. `k_(8)=-16`C. `k_(12)=20`D. `k_(16)=-24` |
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Answer» Correct Answer - B `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` |
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