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

[Calcgeek123]

Homework Statement

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

Homework Equations

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.

 

[Office_Shredder]

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

 

[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?

 

[Office_Shredder]

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

 

[HallsofIvy]

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

 

[Calcgeek123]

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

 

[statdad]

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

 

[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?

 

[Office_Shredder]

they are the same, so the derivative would be zero

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)?

 

[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..

 

[Calcgeek123]

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

 

[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