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

$\lambda =n^{2}$

Hint:

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

Given:

$x=a\: cos\: nt-b\: sin\: nt$

Explanation:

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 $-n^{2}$.