# Proof showing f(x) = ce^x has a constant

1. Sep 20, 2009

### Calcgeek123

1. The problem statement, all variables and given/known data

Show that if f is infinitely differential and f'(x)=f(x), then theres a constant "c" with f(x)=ce^x.

2. Relevant equations

3. The attempt at a solution

I think i'm supposed to begin with an arbitrary function and show that it looks like f(x) = ce^x... but that's as far as I've gotten.

2. Sep 20, 2009

### Office_Shredder

Staff Emeritus
start with c=f(0). Then f(x) = cex is certainly A solution. Suppose g(x) is another solution. What can you conclude about g(x)-cex?

3. Sep 20, 2009

### Calcgeek123

Well.. since f(x)=ce^x is a solution and g(x) is also a soltion, then g(x)-ce^x is the same as g(x)-f(x). Is that right so far?

4. Sep 20, 2009

### Office_Shredder

Staff Emeritus
What can you determine about g(x)-f(x) though?

5. Sep 20, 2009

### HallsofIvy

Staff Emeritus
For example, what is (g(x)- f(x))' ?

6. Sep 20, 2009

### Calcgeek123

(g(x)- f(x))' is the change in slope of graphs g(x) and f(x). Am i getting closer?

7. Sep 20, 2009

Use a rule for simplifying $$(g(x) - f(x))'$$ and then apply what you know about the two functions $$f$$ and $$g$$.

Last edited: Sep 20, 2009
8. Sep 20, 2009

### Calcgeek123

Using the difference rule, if g and f are differentiable functions, then the derivative of the difference of g(x) and f(x) is equal to the derivative of g(x) minus the derivative of f(x).
As for applying what I know about the two functions... they are the same, so the derivative would be zero. Better so far?

9. Sep 20, 2009

### Office_Shredder

Staff Emeritus
Why are they the same? I thought that's what we were trying to prove.

You DO know that
f'(x)=f(x)
g'(x)=g(x)

So (g(x) - f(x))' = g'(x) - f'(x) = ?

Also, what is g(0)-f(0)?

10. Sep 20, 2009

### Calcgeek123

How do i know that a function is equal to its derivative? (referring to f'(x) = f(x), and same with g(x))
Sorry..

11. Sep 20, 2009

### Calcgeek123

Oh, like e^(x) is the derivative of e^(x)?

12. Sep 20, 2009

### Gregg

$$f'(x) = f(x)$$

$$\frac{f'(x)}{f(x)} = 1$$

$$\int \frac{f'(x)}{f(x)}dx = \int dx$$

$$ln|f(x)| = x + C$$

$$f(x) = e^{x+C}$$

$$f(x) = Ae^x$$

where A is just a constant. $$A = \pm e^C$$