The solution of dy/dx + y = e-x, y (0) = 0 is

The correct option is y = xe^-x.

To find the solution to the given first-order linear differential equation, we follow these steps:

Step 1: Identify the type of differential equation
The given equation is:

$$\frac{dy}{dx}+y=e^{-x}$$
This is a standard first-order linear differential equation of the form:

$$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$$
Comparing the two equations, we get:
$P\left(x\right)=1$ and $Q\left(x\right)=e^{-x}$.

Step 2: Find the Integrating Factor (I.F.)
The integrating factor is calculated using the formula:

$$\text{I.F.}=e^{\int P\left(x\right)dx}$$
Substituting $P\left(x\right)=1$:

$$\text{I.F.}=e^{\int 1dx}=e^x$$

Step 3: Find the General Solution
The general solution of a first-order linear differential equation is given by:

$$y\cdot \left(\text{I.F.}\right)=\int Q\left(x\right)\cdot \left(\text{I.F.}\right)dx+C$$
Substitute the values of the Integrating Factor and $Q\left(x\right)$:

$$y\cdot e^x=\int e^{-x}\cdot e^{x}dx+C$$
Since $e^{-x}\cdot e^x=e^0=1$, the equation simplifies to:

$$y\cdot e^x=\int 1dx+C$$
$$y\cdot e^x=x+C$$

Step 4: Apply the initial condition
We are given the initial condition $y\left(0\right)=0$, which means $y=0$ when $x=0$.
Substituting these values into the general solution:

$$0\cdot e^0=0+C$$
$$0=C$$
Thus, the constant of integration $C=0$.

Step 5: Write the final solution
Substitute $C=0$ back into the general equation:

$$y\cdot e^x=x$$
Divide both sides by $e^x$ to express the solution explicitly in terms of $y$:

$$y=\frac{x}{e^x}=x{e}^{-x}$$
Therefore, the particular solution matching the initial condition is indeed y = xe^-x.