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If Y = a E − K T Cos ( P T + C ) , Prove that D 2 Y D T 2 + 2 K D Y D T + N 2 Y = 0 , Where N 2 = P 2 + K 2 ? - Mathematics

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Question

\[\text { If }y = A e^{- kt} \cos\left( pt + c \right), \text { prove that } \frac{d^2 y}{d t^2} + 2k\frac{d y}{d t} + n^2 y = 0, \text { where } n^2 = p^2 + k^2 \] ?

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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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Chapter 12: Higher Order Derivatives - Exercise 12.1 [Page 18]

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RD Sharma Mathematics [English] Class 12
Chapter 12 Higher Order Derivatives
Exercise 12.1 | Q 51 | Page 18

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