If the solve the problem

Question:

If $\mathrm{y}=\mathrm{x}^{\mathrm{n}}\{\mathrm{a} \cos (\log \mathrm{x})+\mathrm{b} \sin (\log \mathrm{x})\}$, prove that $\mathrm{x}^{2} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+(1-2 \mathrm{n}) \frac{\mathrm{dy}}{\mathrm{dx}}+\left(1+\mathrm{n}^{2}\right) \mathrm{y}=0$

Solution:

Formula: -

(i) $\frac{d y}{d x}=y_{1}$ and $\frac{d^{2} y}{d x^{2}}=y_{2}$

(ii) $\frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=\sin \mathrm{x}$

(iii) $\frac{d}{d x} \sin x=-\cos x$

(iv) $\frac{\mathrm{d}(\log \mathrm{x})}{\mathrm{dx}}=\frac{1}{\mathrm{x}}$

(v) $\frac{d}{d x} x^{n}=n x^{n-1}$

(vi) chain rule $\frac{\mathrm{df}}{\mathrm{dx}}=\frac{\mathrm{d}(\text { wou })}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{dw}}{\mathrm{ds}} \cdot \frac{\mathrm{ds}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}}$

Given: -

$y=x^{n}(a \cos (\log x)+b \sin (\log x))$

$\Rightarrow y=a x^{n} \cos (\log x)+b x^{n} \sin (\log x)$

$\frac{d y}{d x}=a n x^{n-1} \cos (\log x)-a x^{n-1} \sin (\log x)+b x^{n-1} \sin \log x+b x^{n-1} \operatorname{coslog} x$

$\Rightarrow \frac{d y}{d x}=x^{n-1} \cos \log x(n a+b)+x^{n-1} \sin (\log x)(b n-a)$

$\Rightarrow \frac{d^{2} y}{d x^{2}}=\frac{d}{d x}\left(x^{n-1} \cos (\log x)(n a+b)+x^{n-1} \sin (\log x)(b n-a)\right)$

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

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

$=x^{n}(n a+b)[(n-1) \cos (\log x)-\sin (\log x)]+(b n-a) x^{n}[(n-1) \sin (\log x)+\cos (\log x)]+(1-2 n) x^{n-}$

${ }^{1} \cos (\log x)(n a+b)+(1-2 n) x^{n-1} \sin (\log x)(b n-a)+a\left(1+n^{2}\right) x^{n} \cos (\log x)+b x^{n}\left(1+n^{2}\right) \sin (\log x)$

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

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