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Beginner question

  1. May 29, 2004 #1
    We just started solving differential equations the other day in my intrductory calculus class. I got every homework question except the last on. here it is.

    [tex] y= ln(3x-2)/3x-2[/tex]

    If someone could get me started that would be great.

    How do I make the division be in numerator-denominator form using the latex code?
     
  2. jcsd
  3. May 29, 2004 #2

    AKG

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    I don't understand the problem. Normally, solving for a differential equation means finding y, some function of x, that satisfies the given equation. If you have y, what is it that you have to do? In case this is worth anything, notice that if:

    [tex]f(x) = ln(3x-2) \mbox{, then } y = \frac{f(x) \times f'(x)}{3}[/tex]

    To get a fraction like this:

    [tex]\frac{dy}{dx}[/tex]

    or this:

    [tex]\frac{ln(3x-2)}{3x-2}[/tex]

    type \frac{dy}{dx} or \frac{ln(3x-2)}{3x-2}.
     
    Last edited: May 29, 2004
  4. May 29, 2004 #3
    Sorry. I tend to get my math vocabulary mixed up. The function above is the derivative and I want to find the original function. Could someone tell how to word this properly.
     
  5. May 29, 2004 #4

    AKG

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    Okay, you're looking for what's called the antiderivative. You can do this by integrating, by performing indefinite integration.

    [tex]\begin{align*}
    \int y dx &= \int \frac{\ln (3x-2)}{3x-2} dx\ \dots \ \mbox{let }u = 3x-2 \\
    &= \frac{1}{3}\int \frac{\ln (u)}{u} du \ \dots \ \mbox{let }v = \ln (u) \\
    &= \frac{1}{3}\int v dv \\
    &= \frac{1}{3} \times \frac{1}{2}v^2 + C \\
    &= \frac{[\ln (3x-2)]^2}{6} + C \\
    \end{align*}[/tex]
     
    Last edited: May 29, 2004
  6. May 29, 2004 #5
    Physics is Phun:

    It sounds like you should have written this to be more clear:

    [tex]y' = \frac{\ln(3x-2)}{3x-2}[/tex]

    The prime (') after the y tells us that this is a differential equation, and our goal is to find y(x).
     
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