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Math Help - Integrating both sides

  1. #1
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    Integrating both sides

    Hi guys, I have a question about whether it's valid to integrate both sides of an equation. The question is asking about the relationship between two functions f,g such that (f/g)'=f'/g'. We're supposed to express f it terms of g and g'. So after doing some work, I got

    \frac{f'}{f}=\frac{(g')^2}{g(g'-g)}

    Now I'd like to integrate both sides to get something like

    \ln f(x)=\int_a^x\frac{(g'(y))^2}{g(y)(g(y)'-g(y))}dy

    which I can then exponentiate to get an expression for f. My question is, is this valid? Are my limits of integration correct? Would I be better off just writing
    \ln \circ f=\int\frac{(g')^2}{g(g'-g)}
    and just keep it all in terms of functions?

    Thanks,
    mtdim
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  2. #2
    Junior Member
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    First of all, there's a small mistake in your work leading to your first displayed line. You'll want to fix that first. Now, you say you want to integrate both sides. To convince yourself this is a valid step, think of it this way: you have an expression of the form Function1 = Function2. We have the theorem:

    Two functions are equal if and only if their antiderivatives differ at most by a constant (provided, of course, that they permit antiderivatives).

    You ask if your limits of integration are correct. Just do this: verify that the integral represents SOME anti-derivative of the original function. (When you take its derivative, do you get back to the original function?) Then the theorem applies. At least, that's how I understand it. I, like you, try to also be explicit about which theorems I'm using for problems like these. That's good. People can be careless about using the FTC.
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  3. #3
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    Thanks for the reply. I think I may just leave it in the second form where there is an implicit constant of integration. That way ALL anti-derivatives are there without having to say something like "for any a...".

    Also, could you point out the mistake you see leading into the first line? I just double checked it and it seems correct (just using the Quotient Rule).
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  4. #4
    Junior Member
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    My bad. I apologize. When I applied the quotient rule, I put g' ^2 in the basement rather than g^2. Not good.

    Yeah, but you're good. Don't forget the absolute value of f in your natural log though. But if you choose to express your answer as an exponential, that becomes no longer necessary. It seems like that's what the problem asks for. so f=Ae^{whatever}
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