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I wasn't really sure how since there aren't any real numbers... only constants. If you know how, here is the solution I found.
And of course I did the work. I aim to master the subject as well as my mind, the best I may.
Last edited by Reuel (2011-01-26 15:51:41)
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Hi;
Might be nothing but there is a difference between your answer and presumably Mathematica'a correct answer.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Which is... ?
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Hi;
Do not panic they are not infallible, just interesting. It gets:
Yours, he thinks can be manipulated to:
It is easy to prove the two forms are not equal by finding one counter example to the premise that they are.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I'll double check my work in the morning. It's been too long a day. Thanks for your input.
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Get some rest, see you tomorrow.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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I woke up early before class and finished the problem. I am confident the solution is correct, even if a typo or two slipped in somewhere, and is here shown. Steps were omitted here and there to reduce length but not so many as to confuse.
The original differential equation:
Separate the variables and integrate both sides to solve the differential equation:
To solve the left integral first, let
The integral for partial fractions is set up by solving for a and b:
The system of equations, then, is
Solving the system yields that
Ergo...
Returning to solve the original differential equation, and that for A(t),
Because B represents area of the growth, it may here be disregarded.
Solving for the constant, C turns out to be 1.
Now, recall that the hyperbolic sine and cosine functions from trigonometry are:
And so, substituting for "x", we get
Factoring gives
Which indeed reduces to
The solution, therefore, is
And there you have it.
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The partial fraction seems okay, have you checked by plugging your answer back into the original DE?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Okay... so... I guess the key is to integrate by partial fractions because it gives a friendler form for answering the question.
You would have
Let
I do not remember how to integrate by partial fractions though. What are the steps for integrating the left side using partial fractions?
Howdy,
Sorry hopefully this isn't too neck-beardish of me but I think the integral was expressed wrong for this.
I think it should be:
I had this problem assigned too what do you think? Obviously this was posted awhile ago, dunno if anyone is still here.
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hi josiahb
Welcome to the forum.
I've only just looked at this problem. It's a long thread and Reuel seems to sort it out in post 32.
That dA should be du by the way.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Oh you're right, he did!! wow I don't know how I missed that, thanks.
Last edited by josiahb (2015-03-09 07:52:08)
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