Question

The differential equation f'(x) = kf(x) has solutions of the form
f(x) = Ce^(kx) (where k and C are constants)

Produce an argument as to why this is the only solution.

This should be an application of the constant Function Theorem.

Answer 1

exponential are only function where derivatives are linear multiples of each other

Answer 2

the constant function theorem is: if f'(x) = 0 for all x in (a,b), then f(x) = constant for all x in (a,b)

Answer 3

how does that show that f(x) = Ce^(kx) are the only solutions? Is there a way to derive this from the theorem?

Answer 4

also you can see when you solve it

f'(x)
---- = k
f(x)

ln(f(x))= kx+ C

exponentiate both side

f(x)= e^kx+C= C e^kx

Answer 5

is there a way I can come to this solution through the constant function theorem?

Answer 6

I don't think constant function theorem address that

Answer 7

it just say if f'(x)=0 then f(x)= some constant

Answer 8

an example of an application of the theorem is:

If f'(x) = g'(x) for all x in (a,b) then g(x) = f(x) + c for some constant c and all x in (a,b)

let h(x) = g(x) - f(x), then h'(x) = g'(x) - f'(x) = 0
by the constant function theorem h(x) = g(x) - f(x) = c
g(x) = f(x) + c

Answer 9

is there something I can do that is similar to this?

Answer 10

what's this constant function theorem, I can't find it anywhere

Answer 11

the constant function theorem is: if f'(x) = 0 for all x in (a,b), then f(x) = constant for all x in (a,b)

Answer 12

that's how I have it defined in my notes

Answer 13

well I can't prove using your theorem , but whenever a function and their derivative set equal with constant multiples, the solution is always in exponential form

Answer 14

how do you know that though?

Answer 15

what's derivative e^x?

Answer 16

what derivative of e^2x

Answer 17

e^x and 2e^2x I think

Answer 18

you see they are constant multiples of each other

Answer 19

why wouldn't the C just be another k then?

Answer 20

C and K are just constant

Answer 21

but they should be the same constant shouldn't they?

Answer 22

2 e^3x ----> 6 e^3x

C=6 k=3

Answer 23

oh alright I can see that now
but, also even if e^x shows this property how do you know it is the only solution?

Answer 24

no other function has that property

Answer 25

but how can you show that this is the only one that works in this situation?

Answer 26

well you gotta solve this differential equation

y=c y'

Answer 27

wouldn't you be solving f'(x) = kf(x)?

Answer 28

ok y(x)=y'(x) k

Answer 29

why is the k on the side with the derivative?

Answer 30

doesn't have to be

k is just a constant

k y(x) =y'(x)

y(x)= 1/k y'(x)

y(x)=k y'(x)

k is just constant , inverse of constant is still a constant

Answer 31

ok I see
how do I solve that equation then?

Answer 32

I will write it as just y from now

k y= y'

k y = dy/dx

ky dx = dy

k dx= 1/y dy

Integrate both sides

kx+C= ln(y)

exponentiate both sides

e^(kx+C)= y

e^C e^kx =y

e^C is just as constant, we might as well right it as C

C e^kx=y

Answer 33

that does make sense now, thanks
I'm still unsure of what my prof wanted me to do with the constant function theorem though. You don't have any more ideas about that do you?

Answer 34

I look that up, didn't find much about it;

but if f'(x)=0 ---> f(x)=C is true

Answer 35

yeah I understand the theorem, I just can't figure out how this problem is an application of it