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

Options

A. `y=e^(x)(x-1)`
B. `y=xe^(-x)`
C. `y=xe^(-x)+1`
D. `y=(x+1)e^(-x)`

Explanation

To solve the differential equation \(\frac{dy}{dx} + y = e^{-x}\) with the initial condition \(y(0) = 0\), we will follow these steps:

### Step 1: Identify the standard form
The given equation is already in the standard linear form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \(P(x) = 1\) and \(Q(x) = e^{-x}\).

### Step 2: Find the integrating factor
The integrating factor \(I(x)\) is given by:
\[ I(x) = e^{\int P(x) \, dx} = e^{\int 1 \, dx} = e^x \]

### Step 3: Multiply the entire equation by the integrating factor
We multiply the original differential equation by \(e^x\):
\[ e^x \frac{dy}{dx} + e^x y = e^x e^{-x} \]
This simplifies to:
\[ e^x \frac{dy}{dx} + e^x y = 1 \]

### Step 4: Rewrite the left side as a derivative
The left side can be rewritten as the derivative of a product:
\[ \frac{d}{dx}(y e^x) = 1 \]

### Step 5: Integrate both sides
Integrating both sides with respect to \(x\):
\[ \int \frac{d}{dx}(y e^x) \, dx = \int 1 \, dx \]
This gives:
\[ y e^x = x + C \]
where \(C\) is the constant of integration.

### Step 6: Solve for \(y\)
Now, we can solve for \(y\):
\[ y = (x + C)e^{-x} \]

### Step 7: Apply the initial condition
Using the initial condition \(y(0) = 0\):
\[ 0 = (0 + C)e^{0} \implies C = 0 \]

### Step 8: Write the final solution
Substituting \(C = 0\) back into the equation for \(y\):
\[ y = x e^{-x} \]

Thus, the solution to the differential equation \(\frac{dy}{dx} + y = e^{-x}\) with the initial condition \(y(0) = 0\) is:
\[ \boxed{y = x e^{-x}} \]