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Provide solution for RD Sharma maths class 12 chapter 11 Higher Order Derivatives exercise very short answer type question 2

Answers (1)


        \lambda =n^{2}


    Double\; di\! f\! \! f\! erentiate\; x\; with\; respect\; to\; t\; to\; get \; \frac{d^{2}x}{dt^{2}},\; then\; substitute\; the\; value\; in\; \frac{d^{2}x}{dt^{2}}=\lambda \; to\; get\; the\; val\! ue\; o\! f\; \lambda .


        x=a\: cos\: nt-b\: sin\: nt


It is given that

        x=a\: cos\: nt-b\: sin\: nt\; \; \; \; \; \; \; \; \; ....(1)

Diff w.r.t t

        \frac{d x}{d t}=a(-\sin n t) n-b n \cos n t \\\\Again\; di\! f\! \! f\; w.r.t \\\\ \frac{d^{2} x}{d t^{2}}=a n(\cos n t)-b n(-\sin n t) \\\\ \frac{d^{2} x}{d t^{2}}=a n^{2} \cos n t+b n^{2} \sin n t \ldots(2) \\\\W\! e \; have, \\\\ \frac{d^{2} x}{d t^{2}}=\lambda x \ldots(3)

Using (1) & (2), (3) becomes

        \begin{aligned} &-a n^{2} \cos n t+b n^{2} \sin n t=\lambda(a \cos n t-b \sin n t) \\ &-n^{2}(a \cos n t+b \sin n t)=\lambda(a \cos n t-b \sin n t) \\ &\therefore \lambda=-n^{2} \end{aligned}

Hence, the value of \lambda  is -n^{2}.

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Gurleen Kaur

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