If the solve the problem


If $y=3 \cos (\log x)+4 \sin (\log x)$, prove that: $x^{2} y_{2}+x y_{1}+y=0$


Note: $y_{2}$ represents second order derivative i.e. $\frac{d^{2} y}{d x^{2}}$ and $y_{1}=d y / d x$


$y=3 \cos (\log x)+4 \sin (\log x) \ldots \ldots$ equation 1

to prove: $x^{2} y_{2}+x y_{1}+y=0$

We notice a second order derivative in the expression to be proved so first take the step to find the second order derivative.

Let's find $\frac{d^{2} y}{d x^{2}}$

As, $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

So, lets first find $d y / d x$

$\frac{d y}{d x}=\frac{d}{d x}(3 \cos (\log x)+4 \sin (\log x))$

Let, $\log x=t$

$\therefore y=3 \cos t+4 \sin t \ldots \ldots \ldots \ldots \ldots$ equation 2

$\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{1}{\mathrm{x}} \ldots \ldots \ldots \ldots .$ equation 3

$\frac{d y}{d t}=-3 \sin t+4 \cos t$

$\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{1}{\mathrm{x}} \ldots \ldots \ldots \ldots$ equation 3

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dt}} \frac{\mathrm{dt}}{\mathrm{dx}}$

$\frac{d y}{d x}=(-3 \sin t+4 \cos t) \frac{1}{x} \ldots \ldots \ldots$ equation 4

Again differentiating w.r.t $\mathrm{x}$ :

Using product rule of differentiation we have

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=(-3 \sin \mathrm{t}+4 \cos \mathrm{t}) \frac{\mathrm{d}}{\mathrm{dx}} \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}} \frac{\mathrm{d}}{\mathrm{dx}}(-3 \sin \mathrm{t}+4 \cos \mathrm{t})$


$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{1}{\mathrm{x}^{2}}(-3 \sin \mathrm{t}+4 \cos \mathrm{t})+\frac{1}{\mathrm{x}} \frac{\mathrm{dt}}{\mathrm{dx}}(-3 \cos \mathrm{t}-4 \sin \mathrm{t})$

Using equation 2,3 and 4 we can substitute above equation as:

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{1}{\mathrm{x}^{2}} \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\frac{1}{\mathrm{x} \mathrm{x}} \frac{1}{(-\mathrm{y})}$

$\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=-\frac{1}{\mathrm{x}} \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{\mathrm{x}^{2}}$

Multiplying $x^{2}$ both sides:

$x^{2} \frac{d^{2} y}{d x^{2}}=-x \frac{d y}{d x}-y$

$\therefore x^{2} y_{2}+x y_{1}+y=0 \ldots \ldots \ldots \ldots$ proved

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