Let `(dy)/(dx) + y = f(x)` where y is a continuous function of x with y(0) = 1 and `f(x) = {{:(e^(-x), if o le x le 2),(e^(-2),if x gt 2):}` Which of the following hold(s) good ?

Options

A. `y(1) = 2e^(-1)`
B. `y'(1) = -e^(-1)`
C. `y(3) = -2e^(-3)`
D. `y'(3) = -2e^(-3)`

Explanation

To solve the differential equation \(\frac{dy}{dx} + y = f(x)\) with the initial condition \(y(0) = 1\) and the piecewise function \(f(x)\), we will follow these steps:

### Step 1: Identify the Integrating Factor
The given differential equation is in the standard form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \(P(x) = 1\) and \(Q(x) = f(x)\). The integrating factor \(I(x)\) is given by:
\[ I(x) = e^{\int P(x) \, dx} = e^{\int 1 \, dx} = e^x \]

**Hint:** The integrating factor is calculated using the exponential of the integral of the coefficient of \(y\).

### Step 2: Multiply the Equation by the Integrating Factor
Multiply the entire differential equation by the integrating factor \(e^x\):
\[ e^x \frac{dy}{dx} + e^x y = e^x f(x) \]

**Hint:** This step transforms the left-hand side into the derivative of a product.

### Step 3: Rewrite the Left-Hand Side
The left-hand side can be rewritten as:
\[ \frac{d}{dx}(e^x y) = e^x f(x) \]

**Hint:** Recognizing the left side as a derivative simplifies the integration process.

### Step 4: Integrate Both Sides
Integrate both sides with respect to \(x\):
\[ \int \frac{d}{dx}(e^x y) \, dx = \int e^x f(x) \, dx \]
This gives:
\[ e^x y = \int e^x f(x) \, dx + C \]

**Hint:** Remember to add the constant of integration \(C\) after integrating.

### Step 5: Solve for \(y\)
Now, solve for \(y\):
\[ y = e^{-x} \left( \int e^x f(x) \, dx + C \right) \]

**Hint:** Isolate \(y\) to express it in terms of \(x\) and the integral.

### Step 6: Evaluate the Integral for \(0 \leq x \leq 2\)
For \(0 \leq x \leq 2\), \(f(x) = e^{-x}\):
\[ \int e^x f(x) \, dx = \int e^x e^{-x} \, dx = \int 1 \, dx = x + C_1 \]
Thus:
\[ y = e^{-x} (x + C_1) \]

**Hint:** The integral simplifies because \(e^x e^{-x} = 1\).

### Step 7: Apply Initial Condition
Using the initial condition \(y(0) = 1\):
\[ 1 = e^0 (0 + C_1) \implies C_1 = 1 \]
So:
\[ y = e^{-x} (x + 1) \]

**Hint:** Substitute the initial condition to find the constant.

### Step 8: Evaluate for \(x > 2\)
For \(x > 2\), \(f(x) = e^{-2}\):
\[ \int e^x f(x) \, dx = \int e^x e^{-2} \, dx = e^{-2} \int e^x \, dx = e^{-2} e^x + C_2 \]
Thus:
\[ y = e^{-x} \left( e^{-2} e^x + C_2 \right) = e^{-2} + C_2 e^{-x} \]

**Hint:** The integral for this range involves \(e^{-2}\) as a constant factor.

### Step 9: Ensure Continuity at \(x = 2\)
To ensure continuity at \(x = 2\):
\[ y(2) = e^{-2} + C_2 e^{-2} \quad \text{and} \quad y(2) = e^{-2} + 2 \]
Setting these equal gives:
\[ e^{-2} + C_2 e^{-2} = e^{-2} + 2 \implies C_2 = 2 \]

**Hint:** Continuity requires that the function values match at the boundary.

### Final Solution
Thus, the complete solution is:
- For \(0 \leq x \leq 2\):
\[ y = e^{-x} (x + 1) \]
- For \(x > 2\):
\[ y = e^{-2} + 2 e^{-x} \]

### Summary of Steps
1. Identify the integrating factor.
2. Multiply the equation by the integrating factor.
3. Rewrite the left-hand side as a derivative.
4. Integrate both sides.
5. Solve for \(y\).
6. Evaluate the integral for the first interval.
7. Apply the initial condition.
8. Evaluate for the second interval.
9. Ensure continuity at the boundary.