Question

Slope Fields/ Separate Variables
Find the equation of the curve, given the derivative and the indicated point on the curve.
dy/dx=lnx/x

Answer 1

point on graph is (1,-2)

Answer 2

@tkhunny

Answer 3

Same deal. Find the general antiderivatve of \(\dfrac{\ln(x)}{x}\).

Answer 4

u mean integral?

Answer 5

Steve.

Answer 6

so is it \[\ln \left| x \right| + C\]

Answer 7

Is that the right function? \(\dfrac{d}{dx}\ln|x| = \dfrac{1}{x}\). Whoops! That's not what we seek.

Answer 8

so thats not the answer?

Answer 9

If you cannot find the derivative and get back to where you started, something went seriously wrong.

\(\int \dfrac{\ln(x)}{x}\;dx = ??\)

Answer 10

so would i use power rule?

Answer 11

You're just guessing, again. Ever hear of "The Substitution Method"?

Answer 12

yes :)

Answer 13

Try u = ln(x) and see where it leads you.

Answer 14

kk

Answer 15

what would be the derivative of \[\frac{ u }{ \ln }\] so i can get du?

Answer 16

Never, EVER write just "ln" again. I would be tempted to fail you for the entire semester for that. "ln" is a function. It REQUIRES and argument.

I am a little concerned. If you are on simple differential equations, you should have a firm grasp on the logarithm function and on integration by substitution.

u = ln(x)

du = (1/x)dx

If you don't know this, you need to go back and find the section where you can re-learn it.

Answer 17

so how did u get du

Answer 18

ok i know what u is but how would i get x?

Answer 19

\(\int \dfrac{\ln(x)}{x}\;dx\)

Given \(u = \ln(x)\;and\;du = \dfrac{dx}{x}\)

This leaves \(\int u\;du\;=\;\dfrac{1}{2}u^{2} + C\)

Finally, \(y = \dfrac{1}{2}(\ln(x))^{2} + C\)

You MUST have seen this, before. Step up your game or you WILL fail this class.