Question

Let \( y(x) \) be the solution of the differential equation \( \frac{dy}{dx} + y = f(x) \), for \( x \geq 0, y(0) = 0 \), where
\[ f(x) = \begin{cases} 2, & 0 \leq x < 1
0, & x \geq 1 \end{cases} \] Then \( y(x) \) is

• A. \( 2(1 - e^{-x}) \) when \( 0 \leq x < 1 \) and \( 2(e^{-x} - 1) \) when \( x \geq 1 \)
• B. \( 2(1 - e^{-x}) \) when \( 0 \leq x < 1 \) and 0 when \( x \geq 1 \)
• C. \( 2(1 - e^{-x}) \) when \( 0 \leq x < 1 \) and \( 2(1 - e^{-1}) e^{-x} \) when \( x \geq 1 \)
• D. \( 2(1 - e^{-x}) \) when \( 0 \leq x < 1 \) and \( 2(1 - e^{-x}) \) when \( x \geq 1 \)

Hint:

When solving linear differential equations with piecewise forcing functions, solve the equation for each piece separately, using the appropriate boundary conditions.

Answer 1

The Correct Option is A

Step 1: Solving the differential equation.
We solve the differential equation \( \frac{dy}{dx} + y = f(x) \) using the method of integrating factors. For \( 0 \leq x < 1 \), the forcing term \( f(x) = 2 \), and for \( x \geq 1 \), \( f(x) = 0 \).

Step 2: Calculating the solution.
For \( 0 \leq x < 1 \), the solution is: \[ y(x) = 2(1 - e^{-x}) \] For \( x \geq 1 \), the solution is: \[ y(x) = 2(e^{-x} - 1) \]

Step 3: Conclusion.
The correct answer is (A) \( 2(1 - e^{-x}) \) when \( 0 \leq x < 1 \) and \( 2(e^{-x} - 1) \) when \( x \geq 1 \).