Question

The solution of the linear differential equation \( \dfrac{dy}{dx}+y=e^{-x} \), when \( x=0,\ y=1 \), is

• A. \( ye^{-x}=x-1 \)
• B. \( ye^{-x}=e^x-1 \)
• C. \( ye^x=x+1 \)
• D. \( ye^{-x}=e^x+1 \)
• E. \( ye^x=x-1 \)

Hint:

For a first-order linear differential equation \( \dfrac{dy}{dx}+P(x)y=Q(x) \), always find the integrating factor \( e^{\int P(x)\,dx} \). It turns the left side into a simple product derivative.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The given equation is a first-order linear differential equation. An equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \) can be solved using an integrating factor.
Step 2: Key Formula or Approach:
1. Identify \( P(x) \) and \( Q(x) \) from the standard form.
2. Calculate the integrating factor (I.F.) using the formula: \( I.F. = e^{\int P(x) dx} \).
3. The general solution is given by: \( y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C \).
4. Use the given initial condition (\( x=0, y=1 \)) to find the value of the constant \( C \).
Step 3: Detailed Explanation:
The given differential equation is \( \frac{dy}{dx} + y = e^{-x} \).
Comparing this with the standard form \( \frac{dy}{dx} + P(x)y = Q(x) \), we have:
\( P(x) = 1 \) and \( Q(x) = e^{-x} \).
Next, we find the integrating factor:
\[ I.F. = e^{\int P(x) dx} = e^{\int 1 dx} = e^x \] The general solution is:
\[ y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C \] \[ y \cdot e^x = \int e^{-x} \cdot e^x dx + C \] The product of the exponentials simplifies: \( e^{-x} \cdot e^x = e^{-x+x} = e^0 = 1 \).
\[ y e^x = \int 1 dx + C \] \[ y e^x = x + C \] This is the general solution. Now we use the initial condition \( y=1 \) when \( x=0 \) to find the particular solution.
Substitute \( x=0 \) and \( y=1 \) into the general solution:
\[ (1) e^0 = 0 + C \] \[ 1 \cdot 1 = C \implies C = 1 \] Substitute the value of \( C \) back into the general solution:
\[ y e^x = x + 1 \] Step 4: Final Answer:
The particular solution of the differential equation is \( ye^x = x+1 \). This corresponds to option (C).