Question

The number of solutions of the differential equation \( \frac{dy}{dx} - y = 1 \), given that \( y(0) = 1 \), is:

• A. 0
• B. 1
• C. 2
• D. infinitely many

Hint:

To solve a first-order linear differential equation, always follow these steps: 1. Write the equation in standard form. 2. Find the integrating factor. 3. Use the integrating factor to simplify and solve the equation. With an initial condition, a \textbf{unique solution} is guaranteed by the existence and uniqueness theorem.

Answer 1

The Correct Option is B

Step 1: Analyze the given differential equation. The equation is: \[ \frac{dy}{dx} - y = 1. \] This is a first-order linear differential equation written in the standard form: \[ \frac{dy}{dx} + P(x)y = Q(x), \] where \( P(x) = -1 \) and \( Q(x) = 1 \). Step 2: Determine the integrating factor. The integrating factor (IF) is given by: \[ \text{IF} = e^{\int P(x) \, dx} = e^{\int -1 \, dx} = e^{-x}. \] Step 3: Multiply through by the integrating factor. Multiplying the entire equation by \( e^{-x} \): \[ e^{-x} \frac{dy}{dx} - e^{-x} y = e^{-x}. \] The left-hand side becomes: \[ \frac{d}{dx}(y \cdot e^{-x}) = e^{-x}. \] Step 4: Solve the simplified equation. Integrate both sides with respect to \( x \): \[ y \cdot e^{-x} = \int e^{-x} \, dx = -e^{-x} + C, \] where \( C \) is the constant of integration. Step 5: Simplify the solution. Multiply through by \( e^x \) to isolate \( y \): \[ y = -1 + Ce^x. \] Step 6: Apply the initial condition \( y(0) = 1 \). Substitute \( x = 0 \) and \( y = 1 \) into the solution: \[ 1 = -1 + C \cdot e^0 \quad \Rightarrow \quad 1 = -1 + C \cdot 1 \quad \Rightarrow \quad C = 2. \] Step 7: Final solution. Substitute \( C = 2 \) into the general solution: \[ y = -1 + 2e^x. \] This is a unique solution that satisfies the initial condition. Thus, the number of solutions is \( \mathbf{1} \).