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Solve this differential equation

  1. Nov 2, 2004 #1
    1. dy/dt + f(t)y = 0

    2. 1/y.dy = -f(t)dt

    3. F(t) = int f(t)dt

    4. int 1/y.dy = - int f(t)dt

    5. ln|y| = -F(t) + A

    I understand up to here, but in my text book the logarithm is eliminated to give:

    6. y = A.e^-F(t)

    I don't understand how you get from step 5. to step 6. Can someone explain it in the simplest possible way (I get confused sometimes when people use shortcuts to explain things, like writing f instead of f(x)).
     
  2. jcsd
  3. Nov 2, 2004 #2

    matt grime

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    2 is dubious and 6 follows by exponentiating both sides, though the A's are different in the two cases:

    if r+k= log(s), then

    exp(r+k)= s (definition of log)

    exp(r)exp(k)= s

    or letting j=exp(k)

    j*exp(r) = s.



    Writing f instead of f(x) isn't a shortcut, it is common practice. Get used to it. That sounds ruder than it is intended, but really, it is very necessary for you to accept and understand that convention. It is also conventional and useful, to write words to explain to yourself, and others what you're doing at each stage in a mathematical argument.
     
    Last edited: Nov 2, 2004
  4. Nov 2, 2004 #3
    it should be (if up to 5 is right) and showing each step

    ln|y| = -F(t) + A
    raise each side to the power of e
    |y|=e^{a-F(t)}
    break the exponential apart across addition
    |y| = e^(a)*e^(-F(t))
    let e^a = b
    |y| = b*e^(-F(t))
     
  5. Nov 2, 2004 #4
    Why is 2. dubious?

    Thank you guys.
     
    Last edited: Nov 2, 2004
  6. Nov 2, 2004 #5

    arildno

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    Dearly Missed

  7. Nov 2, 2004 #6
    I don't really understand your explanation (my maths evidently isn't good enough).

    You were referring to a D.E. [tex]f(y(x))\frac{dy}{dx}=g(x)[/tex], but this is different from the D.E. I gave at the start of this thread.

    Damn, I'm confused.
     
  8. Nov 2, 2004 #7
    ...........................
     
  9. Nov 2, 2004 #8

    arildno

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    No, it is not!
    You have:
    [tex]\frac{1}{y(t)}\frac{dy}{dt}=-f(t)[/tex]
    This has EXACTLY the same form, once you recognize:
    1) "t" is used instead of "x"
    2) -f(t) is used instead of g(x)
    3) [tex]\frac{1}{y(t)}[/tex] is used instead of f(y(x))
     
  10. Nov 2, 2004 #9
    y can't = 0
     
  11. Nov 2, 2004 #10
    Does [tex]f(y(x))\frac{dy}{dx}=g(x)[/tex] mean the derivative of [tex]f(y(x))[/tex] w.r.t x is equal to g(x)?

    I EDITED THIS!
     
  12. Nov 2, 2004 #11
    Is this what the second step should have been?
     
  13. Nov 2, 2004 #12

    arildno

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    No it means that:
    THE PRODUCT OF f, (EVALUATED AT y(x)), WITH THE DERIVATIVE OF y (with respect to x) EQUALS g (i.e., g evaluated at x)
     
  14. Nov 2, 2004 #13
    I thought y is a variable like x, not a function like f.
     
  15. Nov 2, 2004 #14

    arildno

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    This is how it looks like JUST PRIOR to step 2!
     
  16. Nov 3, 2004 #15

    matt grime

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    y is a function of x, otherwise the question would be totally meaningless!##

    "2 is dubious" because it *unnecessarily* treats dx as if it were a number, which at this stage I think I'd prefer you to leave alone until you understand why you can make this abuse of notation. But that's just me.
     
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