Question

Find the differential equation obtained on eliminating c from the equation $y=\left( x+c \right){{e}^{-x}}$.
A. $\dfrac{dy}{dx}-y={{e}^{-x}}$
B. $\dfrac{dy}{dx}-y{{e}^{x}}=1$
C. $\dfrac{dy}{dx}+y{{e}^{x}}=1$
D. $\dfrac{dy}{dx}+y={{e}^{-x}}$

Answer 1

Hint: We will first consider the order of the differential form from the number of unknowns. We will consider the given equation and then keep the term $\left( x+c \right)$ on one side. We will then differentiate both sides with respect to x. We will be using product rules for the same.

Complete step-by-step answer:
The given equation is $y=\left( x+c \right){{e}^{-x}}$.
We know the order of the differential depends on the number of unknown numbers. Order of a differential equation is the order of the highest derivative or differential present in the differential form. An equation needs to get rid of unknowns to become a general or complete solution.
In the given equation $y=\left( x+c \right){{e}^{-x}}$, there is only one unknown c. so, the order will be 1.
On the right side we have multiplication of two x terms. We separate them to keep only $\left( x+c \right)$ as it will help in the differentiation to get rid of c easily.
So, $y=\left( x+c \right){{e}^{-x}}\Rightarrow y{{e}^{x}}=\left( x+c \right)$.
Now we differentiate both sides with respect to x.
$\begin{align}
  & y{{e}^{x}}=\left( x+c \right) \\
 & \Rightarrow \dfrac{d}{dx}\left( y{{e}^{x}} \right)=\dfrac{d}{dx}\left( x+c \right) \\
\end{align}$
We apply the product rule on the LHS and split the terms on the RHS into parts as below,
$\begin{align}
  & \dfrac{d}{dx}\left( y{{e}^{x}} \right)=\dfrac{d}{dx}\left( x+c \right) \\
 & \Rightarrow y\dfrac{d}{dx}\left( {{e}^{x}} \right)+{{e}^{x}}\dfrac{d\left( y \right)}{dx}=\dfrac{d\left( x \right)}{dx}+\dfrac{d\left( c \right)}{dx} \\
 & \Rightarrow y{{e}^{x}}+{{e}^{x}}\dfrac{dy}{dx}=1 \\
\end{align}$
Now we try to match the differential form which is out of the unknown part of the equation with the given options.
We multiply both sides with ${{e}^{-x}}$.
\[\begin{align}
  & y{{e}^{x}}+{{e}^{x}}\dfrac{dy}{dx}=1 \\
 & \Rightarrow {{e}^{-x}}\left( y{{e}^{x}}+{{e}^{x}}\dfrac{dy}{dx} \right)=1.{{e}^{-x}} \\
 & \Rightarrow y+\dfrac{dy}{dx}={{e}^{-x}} \\
 & \Rightarrow \dfrac{dy}{dx}+y={{e}^{-x}} \\
\end{align}\]
The differential form is \[\dfrac{dy}{dx}+y={{e}^{-x}}\]. The correct option is D.

So, the correct answer is “Option D”.

Note: If we keep the form of $\left( x+c \right){{e}^{-x}}$ on the right side and go on differentiating, then we need to find out the value of c from the differential form and replace that in the main equation. Extra steps will be unnecessary as we can avoid that just by breaking the form of the exponential.