Question

The solution of $\left(x+1\right)\frac{dy}{dx}+1=e^{x-y}$ is

• A. $e^y\left(x+1\right)=c$
• B. $e^y\left(x+1\right)=e^x+c$
• C. $e^y\left(x+1\right)=c{e}^x$
• D. None of these

Answer 1

The correct option is B

$e^y\left(x+1\right)=e^x+c$

Explanation of the correct option

The given differential equation: $\left(x+1\right)\frac{dy}{dx}+1=e^{x-y}$.

$$\begin{gathered} \Rightarrow \left(x+1\right)\frac{dy}{dx}+1=\frac{e^x}{e^y} \\ \Rightarrow \left(x+1\right)e^y\frac{dy}{dx}+e^y=e^x \end{gathered}$$

Let us assume that, $\left(x+1\right)e^y=z$

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \Rightarrow \frac{d}{dx}\left[\left(x+1\right)e^y\right]=\frac{dz}{dx} \\ \Rightarrow \left(x+1\right)\frac{d}{dx}\left(e^y\right)+e^y\frac{d}{dx}\left(x+1\right)=\frac{dz}{dx} \\ \Rightarrow \left(x+1\right)e^y\frac{dy}{dx}+e^y=\frac{dz}{dx} \end{gathered}$$

Consider the equation: $\left(x+1\right)e^y\frac{dy}{dx}+e^y=e^x$

$$\begin{gathered} \Rightarrow \frac{dz}{dx}=e^x \\ \Rightarrow dz=e^{x}dx \end{gathered}$$

Integrate both sides of the equation.

$$\begin{gathered} \Rightarrow \int dz=\int e^{x}dx \\ \Rightarrow z=e^x+c\mathbf{[}\mathbf{c}\mathbf{=}\mathbf{Integrating}\mathbf{}\mathbf{Constant}\mathbf{]} \\ \Rightarrow \left(x+1\right)e^y=e^x+c\mathbf{[}\mathbf{\because }\mathbf{z}\mathbf{=}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}{\mathbf{e}}^{\mathbf{y}}\mathbf{]} \end{gathered}$$

Therefore, the solution of the differential equation $\left(x+1\right)\frac{dy}{dx}+1=e^{x-y}$ is $e^y\left(x+1\right)=e^x+c$.

Hence, option B is correct.