Key Concepts and Formulas

• First-Order Linear Differential Equation: A differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$.
• Integrating Factor (IF): For a first-order linear differential equation, the integrating factor is $IF = e^{\int P(x) dx}$.
• General Solution: The general solution to a first-order linear differential equation is given by $y \cdot (IF) = \int Q(x) \cdot (IF) dx + C$, where $C$ is the constant of integration.

Step-by-Step Solution

Step 1: Identify the differential equation and its components.

We are given the differential equation $\frac{dy}{dx} + 2y = f(x)$, which is a first-order linear differential equation. Here, $P(x) = 2$ and $Q(x) = f(x)$.

Step 2: Find the Integrating Factor (IF).

The integrating factor is given by $IF = e^{\int P(x) dx} = e^{\int 2 dx} = e^{2x}$.

Step 3: Solve the differential equation for $x \in [0, 1]$.

In this interval, $f(x) = 1$. Therefore, the differential equation becomes $\frac{dy}{dx} + 2y = 1$. Multiplying both sides by the integrating factor $e^{2x}$, we get: $e^{2x} \frac{dy}{dx} + 2e^{2x}y = e^{2x}$ The left side is the derivative of $y e^{2x}$ with respect to $x$, so we can write: $\frac{d}{dx}(ye^{2x}) = e^{2x}$ Integrating both sides with respect to $x$: $\int \frac{d}{dx}(ye^{2x}) dx = \int e^{2x} dx$ $ye^{2x} = \frac{1}{2}e^{2x} + C_1$ $y = \frac{1}{2} + C_1e^{-2x}$

Step 4: Apply the initial condition $y(0) = 0$ to find $C_1$.

Substituting $x = 0$ and $y = 0$ into the equation, we get: $0 = \frac{1}{2} + C_1e^{0}$ $0 = \frac{1}{2} + C_1$ $C_1 = -\frac{1}{2}$ Therefore, for $x \in [0, 1]$, the solution is: $y(x) = \frac{1}{2} - \frac{1}{2}e^{-2x} = \frac{1}{2}(1 - e^{-2x})$

Step 5: Find $y(1)$.

$y(1) = \frac{1}{2}(1 - e^{-2})$ This value will serve as the initial condition for the next interval.

Step 6: Solve the differential equation for $x > 1$.

In this interval, $f(x) = 0$. Therefore, the differential equation becomes $\frac{dy}{dx} + 2y = 0$. Multiplying both sides by the integrating factor $e^{2x}$, we get: $e^{2x} \frac{dy}{dx} + 2e^{2x}y = 0$ $\frac{d}{dx}(ye^{2x}) = 0$ Integrating both sides with respect to $x$: $\int \frac{d}{dx}(ye^{2x}) dx = \int 0 dx$ $ye^{2x} = C_2$ $y = C_2e^{-2x}$

Step 7: Apply the condition $y(1) = \frac{1}{2}(1 - e^{-2})$ to find $C_2$.

Substituting $x = 1$ and $y = \frac{1}{2}(1 - e^{-2})$ into the equation, we get: $\frac{1}{2}(1 - e^{-2}) = C_2e^{-2}$ $C_2 = \frac{1}{2}e^2(1 - e^{-2}) = \frac{1}{2}(e^2 - 1)$ Therefore, for $x > 1$, the solution is: $y(x) = \frac{1}{2}(e^2 - 1)e^{-2x}$

Step 8: Find $y(\frac{3}{2})$.

Since $\frac{3}{2} > 1$, we use the solution for $x > 1$: $y\left(\frac{3}{2}\right) = \frac{1}{2}(e^2 - 1)e^{-2(\frac{3}{2})} = \frac{1}{2}(e^2 - 1)e^{-3} = \frac{e^2 - 1}{2e^3}$

Common Mistakes & Tips

• Forgetting the Constant of Integration: Always remember to add the constant of integration after performing an indefinite integral.
• Piecewise Functions: When dealing with piecewise functions, solve the differential equation separately for each interval and use the value at the endpoint of one interval as the initial condition for the next.
• Choosing the Correct Solution: Make sure to use the correct solution based on the interval in which you are evaluating the function.

Summary

We solved the given first-order linear differential equation by first finding the integrating factor. Then, we solved the equation separately for the intervals $[0, 1]$ and $(1, \infty)$, using the initial condition $y(0) = 0$ and the continuity of the solution at $x = 1$ to determine the constants of integration. Finally, we evaluated the solution at $x = \frac{3}{2}$.

Final Answer

The final answer is $\boxed{\frac{e^2 - 1}{2e^3}}$, which corresponds to option (D).