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## #1 2006-04-22 09:25:19

Zoom
Guest

### Differential Equation

Ok, so I really need help with this one;

I need to show that y=x * lnx is a solution to the differentialequation y'-(y/x)-1=0

Any help would be really appreciated!

## #2 2006-04-22 17:23:50

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,730

### Re: Differential Equation

Integrating both sides,

logy=log(x-1).

Thats what I get

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## #3 2006-04-22 19:01:55

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### Re: Differential Equation

ganesh wrote:

Integrating both sides,

logy=log(x-1).

Thats what I get

I'm rather unfamiliar with these concepts. (Beginner) What does dx stand for?

## #4 2006-04-22 19:51:19

Zz
Guest

### Re: Differential Equation

And how did you integrate the sides? Is there a formula for this? For future reference.

## #5 2006-04-22 23:33:36

George,Y
Member
Registered: 2006-03-12
Posts: 1,306

instead of

X'(y-Xβ)=0

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## #6 2006-04-23 00:40:54

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Guest

### Re: Differential Equation

What about

y=xlnx
y'=lnx+1

y'-y/x-1 -> lnx+1-xlnx/x-1 =lnx+1-lnx-1=0?

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