Question

If $y = a\cos (\log x) + b\sin (\log x)$ where $a,b$ are parameters then ${x^2}y'' + xy' = $
A. $y$
B. $ - y$
C. $ - 2y$
D. $2y$

Answer 1

Hint: First, we shall analyze the given information so that we can able to solve the problem. Generally in Mathematics, the derivative refers to the rate of change of a function with respect to a variable. First, we need to solve the given equation to obtain$y$. Here, we are applying the product rule and some derivative formulae to find the required answer
 Here we need to find the derivative of $y = a\cos (\log x) + b\sin (\log x)$to obtain the desired answer.
 Formula to be used:
The formulas that are applied in the differentiation of $y = a\cos (\log x) + b\sin (\log x)$are as follows.
$\dfrac{d}{{dx}}\cos x = - \sin x$
$\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
$\dfrac{d}{{dx}}\sin x = \cos x$
$\dfrac{{d\left( {xy} \right)}}{{dx}} = x\dfrac{{dy}}{{dx}} + y$

Complete step by step answer:
It is given that $y = a\cos (\log x) + b\sin (\log x)$ we shall find the derivative of $y$ first. That is $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(a\cos (\log x)) + b\sin (\log x))$
        $ = \dfrac{d}{{dx}}\left( {a\cos (\log x)} \right) + \dfrac{d}{{dx}}\left( {b\sin (\log x)} \right)$
       $ = a\dfrac{d}{{dx}}\cos (\log x) + b\dfrac{d}{{dx}}\sin (\log x)$ ………………$\left( 1 \right)$
     We know that the derivative of $\log x$ is $\dfrac{1}{x}$ , derivative of $\cos x$ is $ - \sin x$ , and the derivative of $\sin x$ is $\cos x$
     That is, $\dfrac{d}{{dx}}\cos x = - \sin x$ ,$\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$ and $\dfrac{d}{{dx}}\sin x = \cos x$ .
Also, $\dfrac{d}{{dx}}\cos (\log x) = - \dfrac{{\sin (\log x)}}{x}$ and $\dfrac{d}{{dx}}\sin (\log x) = \dfrac{{\cos (\log x)}}{x}$
Now, we shall apply the above result in $\left( 1 \right)$
Hence, we get $\dfrac{{dy}}{{dx}} = - \dfrac{{a\sin (\log x)}}{x} + \dfrac{{b\sin (\log x)}}{x}$
$ \Rightarrow x\dfrac{{dy}}{{dx}} = - a\sin (\log x) + b\cos (\log x)$ …….$\left( 2 \right)$
Now, we shall differentiate $\left( 2 \right)$ with respect to $x$
We know that$\dfrac{{d\left( {xy} \right)}}{{dx}} = x\dfrac{{dy}}{{dx}} + y$ .
Thus, we get
\[x\dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{{dy}}{{dx}} = - \dfrac{{a\cos (\log x)}}{x} - \dfrac{{b\sin (\log x)}}{x}\]
$ \Rightarrow x\dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{{dy}}{{dx}} = \dfrac{1}{x}( - a\cos (\log x) - b\sin (\log x))$
$ \Rightarrow {x^2}\dfrac{{{d^2}y}}{{d{x^2}}} + x\dfrac{{dy}}{{dx}} = ( - a\cos (\log x) - b\sin (\log x))$
Since $y = a\cos (\log x) + b\sin (\log x)$ we have
$ \Rightarrow {x^2}\dfrac{{{d^2}y}}{{d{x^2}}} + x\dfrac{{dy}}{{dx}} = - y$……$\left( 3 \right)$
Also,$\dfrac{{dy}}{{dx}}$ can be denoted as $y'$ and $\dfrac{{{d^2}y}}{{d{x^2}}}$ can also be denoted as $y''$ .
Hence $\left( 3 \right)$ becomes ${x^2}y'' + xy' = - y$

So, the correct answer is “Option B”.

Note: If we are asked to calculate the derivative of a given equation, we need to first analyze the given problem where we are able to apply the derivative formulae and the derivative refers to the rate of change of a function with respect to a variable. Also, it is to be noted that$\dfrac{{dy}}{{dx}}$ can be denoted as $y'$ and $\dfrac{{{d^2}y}}{{d{x^2}}}$ can also be denoted as $y''$ .