# (Exponential growth/decay) If f'(x) = -f(x) and f(1) = 1 then

## Homework Statement

If f'(x) = -f(x) and f(1) = 1, then f(x) = ?

(A) $\frac{1}{2}$e-2x+2.
(B) e-x-1.
(C) e1-x.
(D) e-x.
(E) -ex.

## Homework Equations

P(t) = P0ekt.
$\frac{dP}{dt}$ = kP.

## The Attempt at a Solution

I was thinking f(1) = t and 1 = P0, therefore:

P(1) = ek.

Then, I exponetiated both sides and got:

ln(P(1)) = k.

I'm probably going off in a completely different direction, but this is what I did so far.

## Answers and Replies

Dick
Homework Helper
You don't even have to solve the differential equation. You just have to figure out which one of them work. That's the beauty of multiple choice.

I did try that, and the only one I got remotely close to that is letter C, which I got -1 but I set it up as:

-(e1-x)|x = 1 = -1.

The negative is there because -f(x), is that correct or do I leave it out?

EDIT: Oh my, never mind that. I am finding f(1), which if I evaluate letter C at x = 1, I will get 1. Thank you for leading me on the right direction, Dick!

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You are absolutely right, I misread it. Thank you. But say I didn't have the multiple choice, how would I come about doing this problem, or is there no way?

Dick
Homework Helper
You are absolutely right, I misread it. Thank you. But say I didn't have the multiple choice, how would I come about doing this problem, or is there no way?

Sure. Solve the differential equation. df/dt=(-f). Separate the variables, then use the initial condition.

Okay, this is what I did:

$\frac{df}{dt}$ = -f

$\int\frac{df}{-f}$ = $\int dt$

-ln|f| = t

-f = et and obtain:

f = -et.

Where did I go wrong?

Dick
Homework Helper
Okay, this is what I did:

$\frac{df}{dt}$ = -f

$\int\frac{df}{-f}$ = $\int dt$

-ln|f| = t

-f = et and obtain:

f = -et.

Where did I go wrong?

You started out going wrong by forgetting the constant of integration -ln(f)=t+C. Then it got worse when you decided e^(-ln(f))=(-f). That's not right. I'd suggest moving the minus sign over to the t side before you exponentiate.

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Wow, of course. Careless mistake.

Alas! I got it!
Thank you so very much, you were of much help.

Have a good night, my friend.