If 
\[f'(x)=-f(x)\] and 
\[f(1)=2\], then 
\[f(3)=me^n\] for some integers 
\[m\] and 
\[n\].
What are 
\[m\] and 
\[n\]?
\[m=~\]
\[n=~\]


We start with the differential equation given:

\[
f'(x) = -f(x)
\]

This is a first-order linear differential equation. The general solution to this type of equation is:

\[
f(x) = Ce^{-x}
\]

where \( C \) is a constant to be determined.

We are also given the initial condition:

\[
f(1) = 2
\]

Substituting \( x = 1 \) into the general solution:

\[
f(1) = Ce^{-1} = 2
\]

This implies:

\[
C = 2e
\]

Therefore, the specific solution is:

\[
f(x) = 2e \cdot e^{-x} = 2e^{1-x}
\]

Next, we need to find \( f(3) \):

\[
f(3) = 2e^{1-3} = 2e^{-2}
\]

We can express this as:

\[
f(3) = me^n
\]

Comparing, we see that:

\[
m = 2, \quad n = -2
\]

So the values of \( m \) and \( n \) are:

\[
m = 2, \quad n = -2
\]

Would you like more details or have any questions about this solution?

Here are 5 related questions to consider:
1. How would the solution change if \( f'(x) = f(x) \) instead?
2. What is the general solution for \( f'(x) = kf(x) \) where \( k \) is a constant?
3. How does the initial condition affect the solution to a differential equation?
4. What if the initial condition was \( f(0) = 2 \), how would \( f(x) \) be different?
5. Can you solve for \( f(x) \) if given a different form of the differential equation, such as \( f'(x) = -2f(x) \)?

**Tip:** In first-order linear differential equations of the form \( f'(x) = kf(x) \), the solution is typically an exponential function, reflecting the natural growth or decay rate specified by \( k \).

Explore how to solve the differential equation f'(x) = -f(x) with the initial condition f(1) = 2. Understand the application of exponential functions in differential equations. Suitable for advanced high school and college-level students.

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