The differential equation y prime + y = 0 intial condtion is y(0)=2 and the function y(x)=2e^(-x)

To solve the differential equation \( y' + y = 0 \) with the initial condition \( y(0) = 2 \), let's go through the process step by step.

1. **Write the differential equation**: 
   \[
   y' + y = 0
   \]

2. **Rearrange the equation**: 
   \[
   y' = -y
   \]

3. **Separate variables**:
   \[
   \frac{dy}{y} = -dx
   \]

4. **Integrate both sides**:
   \[
   \int \frac{dy}{y} = \int -dx
   \]
   This results in:
   \[
   \ln |y| = -x + C
   \]

5. **Exponentiate to solve for \( y \)**:
   \[
   |y| = e^{-x + C} = e^C e^{-x}
   \]
   Let \( k = e^C \), then:
   \[
   y = k e^{-x}
   \]

6. **Apply the initial condition** \( y(0) = 2 \):
   \[
   2 = k e^{0} \implies k = 2
   \]

7. **Final solution**:
   \[
   y(x) = 2 e^{-x}
   \]

Thus, the solution \( y(x) = 2 e^{-x} \) is indeed correct based on the initial condition provided.

### Questions for Further Exploration:

1. What other initial conditions could be considered for this differential equation?
2. How would the solution change if the equation were \( y' + y = f(x) \) for some function \( f(x) \)?
3. Can you explain the significance of the exponential function in the context of solutions to linear differential equations?
4. What are the long-term behaviors of the solution \( y(x) = 2 e^{-x} \) as \( x \to \infty \)?
5. How can this method of separation of variables be applied to other types of first-order differential equations?

### Tip:
When solving differential equations, always remember to check the initial conditions to ensure your particular solution satisfies the given criteria.

The differential equation y prime + y = 0 initial condition is y(0)=2 and the function y(x)=2e^(-x)

Explore the solution to the differential equation y' + y = 0 with the initial condition y(0)=2, resulting in y(x)=2e^(-x). This detailed guide covers key concepts in differential equations, including separation of variables and exponential functions. Ideal for high school students in Grades 11-12.

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