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Question
\[\text { If }y = A e^{- kt} \cos\left( pt + c \right), \text { prove that } \frac{d^2 y}{d t^2} + 2k\frac{d y}{d t} + n^2 y = 0, \text { where } n^2 = p^2 + k^2 \] ?
Solution
\[\text { We have}, \]
\[y = A e^{- kt} \cos\left( pt + c \right) . . . (1)\]
\[\text { Differentiating y with respect to t, we get }\]
\[\frac{d y}{d t} = - kA e^{- kt} \cos\left( pt + c \right) - pA e^{- kt} \sin\left( pt + c \right)\]
\[ = - ky - pA e^{- kt} \sin\left( pt + c \right) \left[ \text { From } (1) \right]\]
\[ \Rightarrow pA e^{- kt} \sin\left( pt + c \right) = - ky - \frac{d y}{d t} . . . (2)\]
\[\text { Differentiating }\frac{d y}{d t} \text { with respect to t, we get }\]
\[\frac{d^2 y}{d t^2} = - k\frac{d y}{d t} + pkA e^{- kt} \sin\left( pt + c \right) - p^2 A e^{- kt} \cos\left( pt + c \right)\]
\[ = - k\frac{d y}{d t} + k\left( - ky - \frac{d y}{d t} \right) - p^2 y \left[ \text { From }(1) \text { and } \left( 2 \right) \right]\]
\[ = - k\frac{d y}{d t} - k^2 y - k\frac{d y}{d t} - p^2 y\]
\[ = - 2k\frac{d y}{d t} - \left( k^2 + p^2 \right)y\]
\[ \Rightarrow \frac{d^2 y}{d t^2} + 2k\frac{d y}{d t} + \left( k^2 + p^2 \right)y = 0\]
\[ \Rightarrow \frac{d^2 y}{d t^2} + 2k\frac{d y}{d t} + n^2 y = 0, \text { where } n^2 = p^2 + k^2 . \]
\[\text { Hence, }\frac{d^2 y}{d t^2} + 2k\frac{d y}{d t} + n^2 y = 0, \text { where } n^2 = p^2 + k^2 .\]
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