Question

Let \[{{e}^{y}}+xy=e\], then determine the value of the order pair \[\left( \dfrac{dy}{dx},\dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\] at the point \[x=0\].
(a) \[\left( -\dfrac{1}{e},\dfrac{1}{{{e}^{2}}} \right)\]
(b) \[\left( \dfrac{1}{e},\dfrac{1}{{{e}^{2}}} \right)\]
(c) \[\left( \dfrac{1}{e},-\dfrac{1}{{{e}^{2}}} \right)\]
(d) \[\left( -\dfrac{1}{e},-\dfrac{1}{{{e}^{2}}} \right)\]

Answer 1

Hint: In this question, we are given with the equation \[{{e}^{y}}+xy=e\]. In order to determine the value of the order pair \[\left( \dfrac{dy}{dx},\dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\] we will to first differentiate the equation \[{{e}^{y}}+xy=e\] with respect to the variable \[x\]. Now will use the following method for differentiating a function \[f\left( y \right)\] with respect to the variable \[x\]. That is the derivative of the function \[f\left( y \right)\] with respect to the variable \[x\] is given by \[\dfrac{d}{dx}f\left( y \right)=\dfrac{d}{dy}f\left( y \right)\cdot \dfrac{dy}{dx}\].
Then we have to find the value of \[y\] given \[x=0\] and then determine the value of \[\dfrac{dy}{dx}\] at that point. We will then differentiate \[\dfrac{dy}{dx}\] to get the value of \[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}\].

Complete step by step answer:
We are given with the equation \[{{e}^{y}}+xy=e\].
Now in order to determine the value of the order pair \[\left( \dfrac{dy}{dx},\dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\] we will to find the derivative of the equation \[{{e}^{y}}+xy=e\] with respect to the variable \[x\].
Since we know that the derivative of the function \[f\left( y \right)\] with respect to the variable \[x\] is given by \[\dfrac{d}{dx}f\left( y \right)=\dfrac{d}{dy}f\left( y \right)\cdot \dfrac{dy}{dx}\].
Therefore on differentiating the equation \[{{e}^{y}}+xy=e\] with respect to the variable \[x\], we will have
\[{{e}^{y}}\dfrac{dy}{dx}+x\dfrac{dy}{dx}+y=0\]
Now on taking the terms of \[\dfrac{dy}{dx}\] common, we have
\[\left( x+{{e}^{y}} \right)\dfrac{dy}{dx}+y=0\]
Now on substituting the value \[x=0\] in the equation \[{{e}^{y}}+xy=e\], we get
\[\begin{align}
  & {{e}^{y}}+0\times y=e \\
 & \Rightarrow {{e}^{y}}=e \\
 & \Rightarrow y=1
\end{align}\]
Now substituting the values \[x=0\] and \[y=1\] in the differential equation \[\left( x+{{e}^{y}} \right)\dfrac{dy}{dx}+y=0\], we get
\[\begin{align}
  & \left( 0+{{e}^{1}} \right)\dfrac{dy}{dx}+1=0 \\
 & \Rightarrow e\dfrac{dy}{dx}=-1 \\
 & \Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{e}
\end{align}\]
Therefore we have \[\dfrac{dy}{dx}=-\dfrac{1}{e}............(1)\].
Again on differentiating the differential equation \[\left( x+{{e}^{y}} \right)\dfrac{dy}{dx}+y=0\] with respect to the variable \[x\]., we get
\[\begin{align}
  & \dfrac{d}{dx}\left( \left( x+{{e}^{y}} \right)\dfrac{dy}{dx}+y \right)=0 \\
 & \Rightarrow {{e}^{y}}\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\dfrac{dy}{dx}\cdot {{e}^{y}}\cdot \dfrac{dy}{dx}+x\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\dfrac{dy}{dx}+\dfrac{dy}{dx}=0 \\
 & \Rightarrow \left( {{e}^{y}}+x \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \dfrac{dy}{dx} \right)}^{2}}\cdot {{e}^{y}}+2\dfrac{dy}{dx}=0
\end{align}\]
Since we know that \[x+{{e}^{y}}=e\].
Therefore on substituting the value of equation (1) ,\[x+{{e}^{y}}=e\] and \[y=1\] in the above expression, we will have
\[\left( e \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( -\dfrac{1}{e} \right)}^{2}}\cdot {{e}^{1}}+2\left( -\dfrac{1}{e} \right)=0\]
On simplifying the above expression, we get
\[\begin{align}
  & \left( e \right)\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\dfrac{1}{{{e}^{2}}}\cdot {{e}^{1}}-2\left( \dfrac{1}{e} \right)=0 \\
 & \Rightarrow e\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+\dfrac{1}{e}-2\left( \dfrac{1}{e} \right)=0 \\
 & \Rightarrow e\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-\dfrac{1}{e}=0 \\
 & \Rightarrow e\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{e} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{1}{{{e}^{2}}}.....(2)
\end{align}\]
Therefore on combining equation (1) and (2), we get
\[\left( \dfrac{dy}{dx},\dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)=\left( -\dfrac{1}{e},\dfrac{1}{{{e}^{2}}} \right)\]

So, the correct answer is “Option A”.

Note: In this problem, in order to determine the value of the order pair \[\left( \dfrac{dy}{dx},\dfrac{{{d}^{2}}y}{d{{x}^{2}}} \right)\] we will to first differentiate the equation \[{{e}^{y}}+xy=e\] with respect to the variable \[x\]. Now while differentiating the equation \[{{e}^{y}}+xy=e\] with respect to the variable \[x\] take care of the fact that \[y\] is also a function of \[x\].