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

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

Answers (1)

Answer:

\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}

Posted by

Gurleen Kaur

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