Question

Solve: \[ \frac{dy}{dx} + y = e^x, \qquad y(0)=2 \]

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

Hint:

For linear differential equations: \[ \frac{dy}{dx}+Py=Q \] use: \[ I.F.=e^{\int Pdx} \] Then: \[ y(I.F.) = \int Q(I.F.)\,dx + C \] Always apply the initial condition at the end to determine the constant.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The equation \( \frac{dy}{dx} + y = e^x \) is a first-order linear differential equation.
Linear differential equations are equations where the unknown function \( y \) and its derivatives appear only to the first power.
The general form is \( \frac{dy}{dx} + P(x)y = Q(x) \).
The most efficient way to solve this type of equation is the Integrating Factor (I.F.) method.
This method involves finding a factor that, when multiplied by the equation, makes the left side a perfect derivative of a product.
Step 2: Key Formula or Approach:
1. Identify \( P(x) \) and \( Q(x) \).
2. Calculate the Integrating Factor: \( I.F. = e^{\int P(x) dx} \).
3. The solution is given by: \( y \cdot (I.F.) = \int Q(x) \cdot (I.F.) dx + C \).
Step 3: Detailed Explanation:
Comparing \( \frac{dy}{dx} + y = e^x \) with \( \frac{dy}{dx} + P(x)y = Q(x) \):
We find \( P(x) = 1 \) and \( Q(x) = e^x \).
First, calculate the Integrating Factor:
\[ I.F. = e^{\int 1 dx} = e^x \]
Now, multiply the original differential equation by the I.F. (\( e^x \)):
\[ e^x \frac{dy}{dx} + e^x y = e^x \cdot e^x \]
The left-hand side is now the derivative of the product \( y \cdot e^x \):
\[ \frac{d}{dx}(y \cdot e^x) = e^{2x} \]
Integrate both sides with respect to \( x \):
\[ \int \frac{d}{dx}(y e^x) dx = \int e^{2x} dx \]
\[ y e^x = \frac{1}{2} e^{2x} + C \]
To solve for \( y \), divide the entire equation by \( e^x \):
\[ y = \frac{1}{2} \frac{e^{2x}}{e^x} + \frac{C}{e^x} \]
\[ y = \frac{1}{2} e^x + C e^{-x} \]
This is the general solution. Now we use the initial condition \( y(0) = 2 \) to find the specific value of \( C \).
Substitute \( x = 0 \) and \( y = 2 \) into the general solution:
\[ 2 = \frac{1}{2} e^0 + C e^0 \]
Since \( e^0 = 1 \):
\[ 2 = \frac{1}{2} + C \]
\[ C = 2 - \frac{1}{2} = \frac{3}{2} \]
Substituting \( C = \frac{3}{2} \) back into the general solution:
\[ y = \frac{1}{2} e^x + \frac{3}{2} e^{-x} \]
Step 4: Final Answer:
The specific solution to the initial value problem is \( y = \frac{1}{2} e^x + \frac{3}{2} e^{-x} \).
This matches Option (B).