#### Explain solution RD Sharma class 12 chapter Higher Order Derivatives exercise 11.1 question 52 maths

$x^{2} \frac{d^{2} y}{d x^{2}}+x(1-2 n) \frac{d y}{d x}+\left(1+n^{2}\right) y=0$

Hint:

You must know the derivative of

$cos(log\: x)\: and\: sin(log\: x)$

Given:

$\text { If } y=x^{n}\{\operatorname{acos}(\log x)+b \sin (\log x)\} \text { prove that } x^{2} \frac{d^{2} y}{d x^{2}}+x(1-2 n) \frac{d y}{d x}+\left(1+n^{2}\right) y=0$

Solution:

\begin{aligned} &\text { Let } y=x^{n}\{a \cos (\log x)+b \sin (\log x)\}\\ &\text { Use multiplicative rule }\\ &\mathrm{UV}=\mathrm{U}^{\prime} \mathrm{V}+\mathrm{UV}^{\prime}\\ &U=x^{n}\\ &V=a \cos (\log x)+b \sin (\log x) \end{aligned}

\begin{aligned} &\frac{d y}{d x}=n x^{n-1}\{\operatorname{acos}(\log x)+b \sin (\log x)\}+x^{n}\left\{\frac{a \sin (\log x)}{x}+\frac{b \cos (\log x)}{x}\right\} \\ &\frac{d y}{d x}=n \frac{y}{x}+\frac{x^{n}}{x}(-a \sin (\log x)+b \cos (\log x)) \\ &x \frac{d y}{d x}=n y+x^{n}(-a \sin (\log x)+b \cos (\log x)) \end{aligned}

\begin{aligned} &\frac{d y}{d x}+x \frac{d^{2} y}{d x^{2}}=n \frac{d y}{d x}+n x^{n-1}(-a \sin (\log x)+b \cos (\log x))+x^{n}\left\{\frac{a \sin (\log x)}{x}+\frac{b \cos (\log x)}{x}\right\} \\ &\frac{d y}{d x}(1-x)+x \frac{d^{2} y}{d x^{2}}=n x^{n-1}(-a \sin (\log x)+b \cos (\log x))+x^{n}\left\{\frac{a \sin (\log x)}{x}+\frac{b \cos (\log x)}{x}\right\} \end{aligned}

\begin{aligned} &\frac{d y}{d x}(1-n)+x \frac{d^{2} y}{d x^{2}}=n x^{n-1}(-a \sin (\log x)+b \cos (\log x))-\frac{1}{x} y \\ &\frac{d y}{d x}(1-n)+x \frac{d^{2} y}{d x^{2}}=\frac{n}{x}\left(x \frac{d y}{d x}-n y\right)-\frac{y}{x} \end{aligned}

\begin{aligned} &\frac{d y}{d x}+x(1-n) \frac{d y}{d x}=n x \frac{d y}{d x}-n^{2} y-y \\ &\frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}(x-n x-n x)-\left(1+n^{2}\right) y=0 \\ &x^{2} \frac{d^{2} y}{d x^{2}}+x(1-2 n) \frac{d y}{d x}+\left(1+n^{2}\right) y=0 \end{aligned}