Question

Differential equation $\frac{dy}{dx}+y=2,y\left(0\right)=3$

Function y = e^−x + 2

Answer 1

We have,
$y=e^{-x}+2$ ...(1)
Differentiating both sides of (1) with respect to $x$, we get
$$\begin{gathered} \frac{dy}{dx}=-e^{-x} \\ \Rightarrow \frac{dy}{dx}=-\left(y-2\right)\left[\text{Using}\left(1\right)\right] \\ \Rightarrow \frac{dy}{dx}+y=2 \\ \mathrm{It}\text{is the given differential equation.} \end{gathered}$$
$y=e^{-x}+2$ satisfies the given differential equation; hence, it is a solution.
Also, when $x=0,y=e^0+2=1+2=3,\text{i.e.}y\left(0\right)=3.$
Hence, $y=e^{-x}+2$ is the solution to the given initial value problem.