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

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

& {{e}^{y}}+0\times y=e \\\ & \Rightarrow {{e}^{y}}=e \\\ & \Rightarrow y=1 \end{aligned}$$ Now substituting the values $$x=0$$ and $$y=1$$ in the differential equation $$\left( x+{{e}^{y}} \right)\dfrac{dy}{dx}+y=0$$, we get $$\begin{aligned} & \left( 0+{{e}^{1}} \right)\dfrac{dy}{dx}+1=0 \\\ & \Rightarrow e\dfrac{dy}{dx}=-1 \\\ & \Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{e} \end{aligned}$$ Therefore we have $$\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{aligned} & \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{aligned}$$ Since we know that $$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{aligned} & \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{aligned}$$ Therefore on combining equation (1) and (2), we get $$\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$$.