#### Provide solution for RD Sharma maths class 12 chapter 11 Higher Order Derivatives exercise very short answer type question 6

$\frac{d^{2} y}{d x^{2}} \text { in terms of } y \text { is } y$

Hint:

differentiate the equation of $y$  two times to get

$\frac{d^{2}y}{dx}$

& compare its value with $y$

Given:

$y=1-x+\frac{x^{2}}{2 !}+\frac{-x^{3}}{3 !}+\frac{x^{4}}{4 !} \ldots . .\: to\: \infty$

Explanation:

it is given that

$y=1-x+\frac{x^{2}}{2 !}+\frac{-x^{3}}{3 !}+\frac{x^{4}}{4 !} \ldots . \: to\: \infty \; \; \; \; \; \;\; \; \; \ldots\ldots(1)$

Diff $y$ w.r.t to  $x$

\begin{aligned} &\frac{d y}{d x}=-1+\frac{2 x}{2 !}-\frac{-3 x^{2}}{3 !}+\frac{4 x^{3}}{4 !}-\ldots \ldots \text { to } \infty \\&\text { Again diff it w.r.t } x\\ &\begin{aligned} &\frac{d^{2} y}{d x^{2}}=\frac{2}{2 !}-\frac{-6 x^{2}}{3 !}+\frac{4 \cdot 3 \cdot x^{2}}{4 !}-\ldots \ldots \text { To } \infty \\ &\quad=\quad 1-\frac{6 x}{6}+\frac{12 x^{2}}{4 \times 3 \times 2 !}-\ldots . . \text { to } \infty \\&\quad=\quad 1-x+\frac{x^{2}}{2!}-\ldots \text { to } \infty \\ &\quad=\quad \therefore \frac{d^{2} y}{d x}=y \ldots .\; \; \; \; \; \; \; \; (F\! rom\; 1) \end{aligned} \end{aligned}

$Hence,\; \; \frac{d^{2} y}{d x^{2}} = y$

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