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# If $xy=a{{x}^{2}}+\left( \dfrac{b}{x} \right)$, then find the value of $x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=$(A) $\dfrac{y}{x}$ (B) $\dfrac{-y}{x}$(C) $\dfrac{2y}{x}$ (D) $\dfrac{-2y}{x}$

Last updated date: 22nd Mar 2023
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Hint: Carefully examine the given equation and try to convert it such that LHS just has $y$ term. In this way it will be easy for us to find the first and second derivative and substitute it in the final equation to calculate.

The given expression is $xy=a{{x}^{2}}+\left( \dfrac{b}{x} \right)$
We need to find, $x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}.$
For this problem first let us find $\dfrac{dy}{dx}$and $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.
So, consider the given expression,$xy=a{{x}^{2}}+\left( \dfrac{b}{x} \right)$
Dividing with $'x'$ on both sides, we get
$\dfrac{xy}{x}=\dfrac{a{{x}^{2}}}{x}+\dfrac{\dfrac{b}{x}}{x}$
Cancelling the like terms, we get
$\Rightarrow y=ax+\dfrac{b}{{{x}^{2}}}=ax+b{{x}^{-2}}$
Now, by differentiating both sides with respect to$x$, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}(ax)+\dfrac{d}{dx}(b{{x}^{-2}})$
Taking out the constant terms, we get
$\dfrac{dy}{dx}=a\dfrac{d}{dx}(x)+b\dfrac{d}{dx}({{x}^{-2}})$
Now we know, $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n\left( {{x}^{n-1}} \right)$ , so the above equation becomes
\begin{align} & \dfrac{dy}{dx}=a(1)+b\left( -2{{x}^{-2-1}} \right) \\ & \dfrac{dy}{dx}=a+b(-2{{x}^{-3}}) \\ \end{align}
$\dfrac{dy}{dx}=a-2b{{x}^{-3}}............(i)$
Now we find the second derivative.
Let us differentiate both sides of equation (i) with respect to $'x'$, we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( a-2b{{x}^{-3}} \right)$
This can be written as,
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( a \right)-\dfrac{d}{dx}\left( 2b{{x}^{-3}} \right)$
We know differentiation of constant term is zero and taking out the constant terms, we get$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=0-2b\dfrac{d}{dx}\left( {{x}^{-3}} \right)$
Now we know, $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n\left( {{x}^{n-1}} \right)$ , so the above equation becomes
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=-2b\left( -3{{x}^{-3-1}} \right)$
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=6b{{x}^{-4}}.............(ii)$
Now consider the other equation,
$x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=$
Now by substituting values from equation (i) and (ii), we get
$x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=x\left( 6b{{x}^{-4}} \right)+2\left( a-2b{{x}^{-3}} \right)$
\begin{align} & x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=6b{{x}^{-3}}+2a-4b{{x}^{-3}} \\ & x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2a+2b{{x}^{-3}} \\ & x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left( a+\dfrac{b}{{{x}^{3}}} \right) \\ \end{align}
Now by multiplying and dividing with $\left( {{x}^{2}} \right)$ on RHS we get,
\begin{align} & x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left[ a+\dfrac{b}{{{x}^{3}}} \right]\times \dfrac{{{x}^{2}}}{{{x}^{2}}} \\ & x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left[ ax{}^{2}+\dfrac{b{{x}^{2}}}{{{x}^{3}}} \right]\times \dfrac{1}{{{x}^{2}}} \\ & x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left[ a{{x}^{2}}+\dfrac{b}{x} \right]\dfrac{1}{{{x}^{2}}} \\ \end{align}
But given, $xy=a{{x}^{2}}+\left( \dfrac{b}{x} \right)$, so the above equation becomes,
$x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\left( xy \right)\dfrac{1}{{{x}^{2}}}$
Cancelling the like terms, we get
$x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+2\dfrac{dy}{dx}=2\dfrac{y}{x}$
Hence the correct answer is option (C).
Note: Another way to solve this problem is directly differentiating the given expression with respect to $x$ instead of dividing the expression by $x$ .
$\dfrac{d(xy)}{dx}=\dfrac{d}{dx}\left( a{{x}^{2}}+\left( \dfrac{b}{x} \right) \right)$
Now we will apply product rule, so we get
$x\dfrac{d(y)}{dx}+y\dfrac{d(x)}{dy}=\dfrac{d}{dx}\left( a{{x}^{2}} \right)+\dfrac{d}{dx}\left( \dfrac{b}{x} \right)$
This gets a little complicated. But you will get the same answer.