#### Need solution for RD Sharma maths class 12 chapter Higher Order Derivatives exercise 11.1 question 51

$\frac{d^{2} y}{d t^{2}}+2 k \frac{d y}{d t}+n^{2} y=0$

Hint:

You must know the derivative of

$A e^{-k t} \cos (p t+c)$

Given:

$\text { If } y=A e^{k t} \cos (p t+c) \text { prove that } \frac{d^{2} y}{d t^{2}}+2 k \frac{d y}{d t}+n^{2} y=0 \text { where } n^{2}=p^{2}+k^{2}$

Solution:

\begin{aligned} &\text { Let } y=A e^{-k t} \cos (p t+c) \quad \ldots . .(1) \\ &\frac{d y}{d x}=A e^{-k t}(-k) \cos (p t+c)+A e^{-k t}(-\sin (p t+c) p) \\ &\frac{d y}{d t}=-A k e^{-k t} \cos (p t+c)-A p e^{-k t}(\sin (p t+c)) \quad \ldots . .(2) \end{aligned}

\begin{aligned} &\frac{d^{2} y}{d x^{2}}=A k^{2} e^{-k t} \cos (p t+c)+A p k e^{-k t} \sin (p t+c)-A p k e^{-k t} \sin (p t+c)-A p^{2} e^{-k t} \cos (p t+c) \\ &\frac{d^{2} y}{d t^{2}}=\left(k^{2}-p^{2}\right) A e^{-k t} \cos (p t+c)+2 k\left(A p e^{-k t} \sin (p t+c)\right) \quad \ldots . .(3) \end{aligned}

\begin{aligned} &\text { Using }(1)\: \&\: (2)\: \&\: n^{2}=k^{2}+p^{2} \\ &\frac{d^{2} y}{d t^{2}}=\left(k^{2}-p^{2}\right) y+2 k\left(-k y-\frac{d y}{d t}\right) \\ &\frac{d^{2} y}{d t^{2}}=\left(k^{2}-p^{2}-2 k^{2}\right) y+2 k\left(\frac{d y}{d t}\right) \\ &\frac{d^{2} y}{d t^{2}}=-n^{2} y-2 k \frac{d y}{d t} \\ &\frac{d^{2} y}{d t^{2}}+2 k \frac{d y}{d t}+n^{2} y=0 \end{aligned}

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