Question

If $x+y={{\tan }^{-1}}y\,and\,{{y}^{''}}=f\left( y \right){{y}^{'}}\,then\,f\left( y \right)=$
(a) $\dfrac{1}{y\left( 1+{{y}^{2}} \right)}$
(b) $\dfrac{3}{y\left( 1+{{y}^{2}} \right)}$
(c) $\dfrac{2}{y\left( 1+{{y}^{2}} \right)}$
(d) $\dfrac{-2}{y\left( 1+{{y}^{2}} \right)}$

Answer 1

Hint: We start solving the problem by differentiating both LHS and RHS with respect to x. We use the properties $\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)$ and $\dfrac{d}{dx}\left( f\left( y \right) \right)=\dfrac{d}{dy}f\left( y \right)\times \dfrac{dy}{dx}$ to proceed further through the problem. We then use \[\dfrac{d\left( {{\tan }^{-1}}y \right)}{dy}=\dfrac{1}{1+{{y}^{2}}}\] and make necessary calculations and arrangements in the problem to get the desired result.

Complete step by step answer:
Given equation: $x+y={{\tan }^{-1}}y........(i)$
On differentiating both sides with respect to x-
$\dfrac{d\left( x+y \right)}{dx}=\dfrac{d\left( {{\tan }^{-1}}y \right)}{dy}............(ii)$\[\]
Important property of differentiation-
a. $\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)$.
b. $\dfrac{d}{dx}\left( f\left( y \right) \right)=\dfrac{d}{dy}f\left( y \right)\times \dfrac{dy}{dx}$.
By using the property $\dfrac{d}{dx}\left( f\left( x \right)+g\left( x \right) \right)=\dfrac{d}{dx}f\left( x \right)+\dfrac{d}{dx}g\left( x \right)$ in equation (ii) -
\[\dfrac{dx}{dx}+\dfrac{dy}{dx}=\dfrac{d\left( {{\tan }^{-1}}y \right)}{dx}.............(iii)\]
We will use standard notation of differentiation:
$\begin{align}
  & y'=\dfrac{dy}{dx}\,and \\
 & y''=\dfrac{{{d}^{2}}y}{d{{x}^{2}}} \\
\end{align}$
Now, further solving equation (iii), we get :
$1+y'=\dfrac{d}{dx}\left( {{\tan }^{-1}}y \right)$
By using property $\dfrac{d}{dx}\left( fy \right)=\dfrac{d}{dy}fy\times \dfrac{dy}{dx}$ , we get
\[\begin{align}
  & 1+{y’}=\dfrac{d}{dy}{{\tan }^{-1}}y\times \dfrac{dy}{dx} \\
 & \Rightarrow 1+{y’}=\dfrac{d}{dy}\left( {{\tan }^{-1}}y \right)\times y' \\
\end{align}\]
We know that differentiation of \[{{\tan }^{-1}}y\] w.r.t \[y=\dfrac{1}{1+{{y}^{2}}}\]
i.e. \[\dfrac{d\left( {{\tan }^{-1}}y \right)}{dy}=\dfrac{1}{1+{{y}^{2}}}\]
Using this in above equation-
\[1+y'=y'\times \left( \dfrac{1}{1+{{y}^{2}}} \right)............(iv)\]
Now, on again differentiating both sides with respect to x, we will get-
\[\dfrac{d}{dx}\left( 1+y' \right)=\dfrac{d}{dx}\left( \dfrac{y'}{1+{{y}^{2}}} \right)............(v)\]
\[\dfrac{d}{dx}\left( 1 \right)+\dfrac{d}{dx}\left( y' \right)=\dfrac{d}{dx}\left\{ y'\times \dfrac{1}{1+{{y}^{2}}} \right\}\]
\[\begin{align}
  & \Rightarrow 0+\dfrac{d}{dx}\left( y' \right)=\dfrac{d}{dx}\left\{ y'\times \dfrac{1}{1+{{y}^{2}}} \right\} \\
 & \\
\end{align}\]
Using standard notation conversation
\[\begin{align}
  & \dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{d}{dx}\left\{ y'\times \dfrac{1}{1+{{y}^{2}}} \right\} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left\{ y'\times \dfrac{1}{1+{{y}^{2}}} \right\} \\
 & \Rightarrow y''=\dfrac{d}{dx}\left\{ y'\times \dfrac{1}{1+{{y}^{2}}} \right\} \\
\end{align}\]
Now, we know that the product rule of differentiation is given as :\[\dfrac{d}{dx}\left\{ f\left( x \right)\times g\left( x \right) \right\}=f\left( x \right)\times \dfrac{d}{dx}g\left( x \right)+g\left( x \right)\times \dfrac{d}{dx}f\left( x \right)\]
\[\Rightarrow y''=y'\times \dfrac{d}{dx}\left( \dfrac{1}{1+{{y}^{2}}} \right)+\dfrac{1}{1+{{y}^{2}}}\times \dfrac{d}{dx}\left( y' \right)\]
\[\Rightarrow y''=y'\times \dfrac{d}{dy}\left( \dfrac{1}{1+{{y}^{2}}} \right)\times \dfrac{dy}{dx}+\dfrac{1}{1+{{y}^{2}}}\times \dfrac{d}{dx}\left( y' \right)\]
We know, \[\dfrac{d}{dy}\left( \dfrac{1}{1+{{y}^{2}}} \right)=\dfrac{d}{dy}{{\left( 1+{{y}^{2}} \right)}^{-1}}=-\left( 2y \right){{\left( 1+{{y}^{2}} \right)}^{-2}}=\dfrac{-1\left( 2y \right)}{{{\left( 1+{{y}^{2}} \right)}^{2}}}\] .
Putting this value in the above equation, we will get.
$\Rightarrow y''=y'\times \left( \dfrac{-2y}{{{\left( 1+{{y}^{2}} \right)}^{2}}} \right)+\dfrac{1}{1+{{y}^{2}}}\dfrac{d}{dx}\left( y' \right)$
We know $\begin{align}
  & \dfrac{d}{dx}\left( y' \right)=\dfrac{d}{dx}\left( \dfrac{dy}{dx} \right)=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=y'' \\
 & \\
\end{align}$
$\begin{align}
  & \Rightarrow y''=y'\left( \dfrac{-2y}{{{\left( 1+{{y}^{2}} \right)}^{2}}} \right)+\dfrac{1}{1+{{y}^{2}}}y'' \\
 & \\
\end{align}$
Taking all the terms containing y’’ to one side of equation, we will get-
$\begin{align}
  & \Rightarrow y''\left( 1-\dfrac{1}{1+{{y}^{2}}} \right)=y'\left( \dfrac{-2y}{{{\left( 1+{{y}^{2}} \right)}^{2}}} \right) \\
 & \Rightarrow y''\left( \dfrac{1+{{y}^{2}}-1}{1+{{y}^{2}}} \right)=y'\left( \dfrac{-2y}{{{\left( 1+{{y}^{2}} \right)}^{2}}} \right) \\
\end{align}$
Dividing both sides of equation by y, we will get,
 $\Rightarrow y''\left[ \dfrac{y}{{{y}^{2}}+1} \right]=\dfrac{-2y'}{{{\left( 1+{{y}^{2}} \right)}^{2}}}$
Multiplying both sides of equation by $\left( 1+{{y}^{2}} \right)$ we will get-
$\begin{align}
  & \Rightarrow y\times y''=\dfrac{-2y'}{{{\left( 1+{{y}^{2}} \right)}}} \\
 & \Rightarrow y''=\dfrac{-2}{y{{\left( 1+{{y}^{2}} \right)}}}y' \\
\end{align}$
Given $y''=f\left( y \right)y'$, so $f\left( y \right)=\dfrac{-2}{y{{\left( 1+{{y}^{2}} \right)}}}\,$

So, the correct answer is “Option D”.

Note: We should not make mistakes in signs, additions while performing differentiation operations. We can also solve the problem by differentiating both sides with respect to y for the first derivative to get $\dfrac{dx}{dy}$ and then take the inverse of it to get the equation for $y'$. We can then differentiate $y'$ in order to get $y''$. After finding both $y'$ and $y''$, we can compare both to get the required result.