Question

The general solution of the differential equation \( \frac{dy}{dx} + 2y = e^{-x} \) is:

• A. \( y = e^{-x} + Ce^{-2x} \)
• B. \( y = \frac{1}{3} e^{-x} + Ce^{-2x} \)
• C. \( y = e^{-x} + Ce^{2x} \)
• D. \( y = \frac{1}{2} e^{-x} + Ce^{2x} \)

Hint:

Key Fact: For first-order linear differential equations, use the integrating factor to simplify the solution process.

Answer 1

The Correct Option is B

Step 1: Identify the Equation Type
This is a first-order linear differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \), with \( P(x) = 2 \), \( Q(x) = e^{-x} \).

Step 2: Find Integrating Factor (I.F.)
\[ \text{I.F.} = e^{\int 2 \, dx} = e^{2x} \]

Step 3: Multiply through by I.F. and Integrate
\[ e^{2x} \frac{dy}{dx} + 2e^{2x} y = e^{x} \Rightarrow \frac{d}{dx}(y e^{2x}) = e^{x} \] \[ \Rightarrow y e^{2x} = \int e^{x} dx = e^{x} + C \Rightarrow y = e^{-x} + C e^{-2x} \]

Alternate Method: Using Homogeneous + Particular Solution
Homogeneous: \( y_h = C e^{-2x} \) 
Try particular solution \( y_p = a e^{-x} \) 
Substitute: \( -a e^{-x} + 2a e^{-x} = e^{-x} \Rightarrow a = 1 \) 
General solution: \( y = e^{-x} + C e^{-2x} \)