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**Hint:**Here from the given equation, first of all we will be differentiating the above equation with respect to x which will give us the first order derivative of the above equation that will be termed as $\dfrac{dy}{dx}$. We will be using the product rule which is given by $\dfrac{d}{dx}\left[ f\left( x \right)g\left( x \right) \right]=g(x)f'(x)-f(x)g'(x)$ for the second derivative that is termed as $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$ .On substituting these values in the show that equation, we will be getting the left hand side to be equal to right hand side.

**Complete step by step answer:**

Now, the given equation is $y=3\cos \left( \log x \right)+4\sin (\log x)$

We know that the chain rule of the equation for the two functions $f(x)$ and $g(x)$ is given by

$\dfrac{d}{dx}\left[ f(g(x)) \right]=f'(g(x))g'(x)$

We know that the general derivative of the function $\cos x$ is given by $-\sin x$ and $\log x$ is given by $\dfrac{1}{x}$.

On differentiating the above equation and applying chain rule with respect to $x$ , we get

$\dfrac{dy}{dx}=-\dfrac{3\sin \left( \log x \right)}{x}+\dfrac{4\cos (\log x)}{x}$

Which can be further expressed as

${{y}_{1}}=\dfrac{1}{x}\left[ -3\sin \left( \log x \right)+4\cos (\log x) \right]......\left( 1 \right)$

We know that the product rule of derivative is given by $\dfrac{d}{dx}\left[ f\left( x \right)g\left( x \right) \right]=g(x)f'(x)-f(x)g'(x)$

Where $f(x)$ is taken as $\dfrac{1}{x}$ and $g(x)$ is taken as $\left[ -3\sin \left( \log x \right)+4\cos (\log x) \right]$.

So, on further differentiating equation $\left( 1 \right)$ and applying chain rule, we get the equation as

Again, further differentiating, we get the equation as

$\begin{align}

& \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d\left( \dfrac{1}{x} \right)}{dx}\left( -3\sin \left( \log x \right)+4\cos \left( \log x \right) \right)+\dfrac{1}{x}\left( \dfrac{d\left( -3\sin \left( \log x \right)+4\cos \left( \log x \right) \right)}{dx} \right) \\

& \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-\left[ -3\sin \left( \log x \right)+4\cos \left( \log x \right) \right]}{{{x}^{2}}}+\dfrac{1}{x}\left[ -3\cos \left( \log x \right)\dfrac{1}{x}-4\sin \left( \log x \right)\dfrac{1}{x} \right] \\

& \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{3\sin \left( \log x \right)-4\cos \left( \log x \right)-3\cos \left( \log x \right)-4\sin \left( \log x \right)}{{{x}^{2}}} \\

\end{align}$

Now reducing it further the equation becomes

$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-\sin (\log x)-7\cos (\log x)}{{{x}^{2}}}$

Which can further be expressed as

${{y}_{2}}=\dfrac{-\sin (\log x)-7\cos (\log x)}{{{x}^{2}}}$

Now the equation of LHS side can be written in short form as $LHS={{x}^{2}}{{y}_{2}}+x{{y}_{1}}+y$

Now substituting all the values, the equation becomes

$LHS={{x}^{2}}\left( \dfrac{-\sin (\log x)-7\cos (\log x)}{{{x}^{2}}} \right)+x\left( -\dfrac{3\sin \left( \log x \right)}{x}+\dfrac{4\cos (\log x)}{x} \right)+3\cos \left( \log x \right)+4\sin (\log x)$Multiplying the terms by opening the bracket, we get the equation as

$LHS=-\sin (\log x)-7\cos (\log x)-3\sin (\log x)+4\cos (\log x)+3\cos (\log x)+4sin\left( \log x \right)$

which can further be rearranged into

All the terms cancel out each other and we get a 0.

Hence LHS becomes equal to RHS.

Hence the solution is justified.

**Note:**

In this question we must be cautious in finding the derivatives and the student must thoroughly know the derivatives of the functions such as $\cos ,\sin and\log $ functions. Further the student must be very careful in taking out the common terms to make the left hand side equal to zero. Also be careful with the signs used as mistakes can be committed if improper sign creeps in due to silly mistakes.

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