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

[Alcubierre]

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) e^1-x.
(D) e^-x.
(E) -e^x.

Homework Equations

P(t) = P₀e^kt.
$\frac{dP}{dt}$ = kP.

The Attempt at a Solution

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

P(1) = e^k.

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.

 

[Dick]

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.

 

[Alcubierre]

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:

-(e^1-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!

 

[Alcubierre]

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]

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.

 

[Alcubierre]

Okay, this is what I did:

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

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

-ln|f| = t

-f = e^t and obtain:

f = -e^t.

Where did I go wrong?

 

[Dick]

Okay, this is what I did:

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

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

-ln|f| = t

-f = e^t and obtain:

f = -e^t.

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.

 

[Alcubierre]

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.