Question

What is the differential equation for $\frac{dy}{dx}=x{e}^{\left(y-2x\right)}$?

Answer 1

Solution.

The differential equation is

$$\begin{gathered} \frac{dy}{dx}=x{e}^{\left(y-2x\right)} \\ \Rightarrow \frac{dy}{dx}=x{e}^y{e}^{-2x} \\ \Rightarrow \frac{dy}{e^y}=x{e}^{-2x}dx \end{gathered}$$

Integrating both sides

$$\begin{gathered} \int \frac{dy}{e^y}=\int x{e}^{-2x}dx \\ \Rightarrow \int e^{-y}dy=\int x{e}^{-2x}dx \\ \Rightarrow -e^{-y}=x\left[-\frac{1}{2}{e}^{-2x}\right]-\int 1\left[-\frac{1}{2}{e}^{-2x}\right]dx \\ \Rightarrow -e^{-y}=-\frac{1}{2}x{e}^{-2x}-\left[\frac{1}{4}{e}^{-2x}\right]+c \\ \therefore e^{-y}=-\frac{1}{2}x{e}^{-2x}\left[x+\frac{1}{2}\right]+c \end{gathered}$$

Hence, the solution to the given differential equation is $e^{-y}=-\frac{1}{2}x{e}^{-2x}\left[x+\frac{1}{2}\right]+c$.