- #1

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[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?

- Thread starter Physics is Phun
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- #1

- 95

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[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

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:Physics is Phun said:

[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?

[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:

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- #4

AKG

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Okay, you're looking for what's called thePhysics is Phun said:

[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:

- #5

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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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