 # Find the solution of the following differential equations. dy=e^{x-y}dx abreviatsjw 2021-12-28 Answered
Find the solution of the following differential equations.
$dy={e}^{x-y}dx$
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Step 1
$dy={e}^{x-y}dx$
$dy={e}^{x}\cdot {e}^{-y}dx$
$\frac{dy}{{e}^{-y}}={e}^{x}dz$
Step 2
${e}^{y}dy={e}^{x}dx$
Integrting both side
$\int {e}^{y}dy=\int {e}^{x}dx$.
${e}^{y}={e}^{x}+c$
###### Not exactly what you’re looking for? Jimmy Macias
In this tutorial we shall evaluate the simple differential equation of the form $\frac{dy}{dx}={e}^{x-y}$ using the method of separating the variables.
The differential equation of the form is given as
$\frac{dy}{dx}={e}^{x-y}$
$⇒\frac{dy}{dx}={e}^{x}{e}^{-y}$
$⇒\frac{dy}{dx}=\frac{{e}^{x}}{{e}^{y}}$
Separating the variables, the given differential equation can be written as
${e}^{y}dy={e}^{x}dx$ (i)
In the separating the variables technique we must keep the terms dy and dx in the numerators with their respective functions.
Now integrating both sides of the equation (i), we have
$\int {e}^{y}dy=\int {e}^{x}dx$
Using the formulas of integration $\int {e}^{x}dx={e}^{x}$, we get
${e}^{y}={e}^{x}+c$
$⇒y=\mathrm{ln}\left({e}^{x}+c\right)$
This is the required solution of the given differential equation.
###### Not exactly what you’re looking for? Vasquez

$\begin{array}{}dy={e}^{x-y}dx\\ dy={e}^{x}\cdot {e}^{-y}dx\\ \frac{dy}{{e}^{-y}}={e}^{x}dz\\ {e}^{y}dy={e}^{x}dx\\ \int {e}^{y}dy=\int {e}^{x}dx.\\ {e}^{y}={e}^{x}+c\end{array}$