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**Zoom****Guest**

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!

**ganesh****Administrator**- Registered: 2005-06-28
- Posts: 23,437

Integrating both sides,

logy=log(x-1).

Thats what I get

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**Zoom****Guest**

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?

**Zz****Guest**

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

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

instead of

**X'(y-Xβ)=0**

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**Zoom****Guest**

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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