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

Correct Answer:
The correct option is y = xe^-x.

Step-by-Step Explanation:

We are given the first-order linear differential equation:
$$\frac{dy}{dx}+y=e^{-x}$$ with the initial condition $y\left(0\right)=0$.

Step 1: Identify the type of differential equation
A first-order linear differential equation is of the standard form:
$$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$$ Comparing this with our given equation, we have:
$P\left(x\right)=1$
$Q\left(x\right)=e^{-x}$

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: Multiply the differential equation by the Integrating Factor
Multiplying both sides of the original differential equation by $e^x$ gives:
$$e^x\frac{dy}{dx}+e^{x}y=e^x\cdot e^{-x}$$ By the product rule of differentiation, the left-hand side can be written as the derivative of the product of $y$ and the integrating factor:
$$\frac{d}{dx}(y\cdot e^x)=1$$

Step 4: Integrate both sides with respect to x
Integrating both sides gives:
$$y\cdot e^x=\int 1dx$$ $$y\cdot e^x=x+C$$ where $C$ is the constant of integration.

Dividing both sides by $e^x$ (or multiplying by $e^{-x}$), we get the general solution:
$$y=(x+C)e^{-x}$$

Step 5: Apply the initial condition to find C
We are given that $y\left(0\right)=0$, which means when $x=0$, $y=0$.
Substituting these values into the general solution equation:
$$0=(0+C)e^0$$ Since $e^0=1$:
$$0=C\cdot 1\Rightarrow C=0$$

Step 6: Write the final particular solution
Substituting $C=0$ back into the general solution equation:
$$y=x{e}^{-x}$$