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

  • Thread starter Alcubierre
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Homework Statement



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

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

Homework Equations



P(t) = P0ekt.
[itex]\frac{dP}{dt}[/itex] = 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

  • #2
Dick
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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.
 
  • #3
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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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  • #4
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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?
 
  • #5
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?
Sure. Solve the differential equation. df/dt=(-f). Separate the variables, then use the initial condition.
 
  • #6
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Okay, this is what I did:

[itex]\frac{df}{dt}[/itex] = -f

[itex]\int\frac{df}{-f}[/itex] = [itex]\int dt[/itex]

-ln|f| = t

-f = et and obtain:

f = -et.

Where did I go wrong?
 
  • #7
Dick
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Okay, this is what I did:

[itex]\frac{df}{dt}[/itex] = -f

[itex]\int\frac{df}{-f}[/itex] = [itex]\int dt[/itex]

-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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  • #8
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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.
 

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