Finding the intergral function (dQ/dt) = kQ

  • Thread starter miniradman
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  • #1
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Homework Statement


Find the integral (I think I'm finding the integral) of:
[itex]\frac{dQ}{dt}[/itex] =kQ

Homework Equations





The Attempt at a Solution


I just got feedback from my teacher, and he told me I make a mistake somewhere in this question. But I don't know where I've gone wrong?

[itex]\frac{dQ}{dt}[/itex]=kQ
[itex]\frac{dQ}{Q}[/itex]=kdt
∫[itex]\frac{dQ}{Q}[/itex]=∫kdt
lnQ=kt+c
Q=ekt+ec
Where ec is a constant, so let ec=A
Q=Aekt

∴ Q=Aekt

Can anyone see where I've gone wrong?
 
Last edited:

Answers and Replies

  • #2
1,344
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If you substitute your solution back into the differential equation, the LHS equals the RHS, so the solution is fine. However, you wrote in an intermediate calculation that Q = ekt + ec. It can't be an addition. Try and think why it can't be an addition.

Hint: What are rules for algebraic operations with indices?

You made the same mistake in your next post.
 
  • #3
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oh... Kt + c will have to go to

ekt+c
ektx ec

right?
 
  • #4
1,344
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Yup, I can't see any other mistakes. This answer should score full marks now.
 
  • #5
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awesome!

Thanks man, I owe you one
 
  • #6
HallsofIvy
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Technically, there is one other error. The integral, [itex]\int (1/Q)dQ= ln|Q|[/itex], not ln(Q).
 
  • #7
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ahhh... ok, thanks for that
 
  • #8
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Minor point: there is no such word as "intergral" in the English language. Seeing as you have started two threads with this in the title, I thought I should point it out.
 
  • #9
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Sorry, I have terrible spelling :frown:
 

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