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 March 10th, 2011, 01:59 PM #21 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Exact form, differential equations oh yes, i multiplied x by the original equation , i dont know why! integrating, xy=ln(x)+A y= (ln(x)+A) / x
 March 10th, 2011, 02:07 PM #22 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Exact form, differential equations While there is nothing wrong with the solution as you gave it, you can write in in a simpler form: $y=\frac{1}{x}$$\ln(x)+A$$$ Since A is an arbitrary constant, we may write it as ln(A): $y=\frac{1}{x}$$\ln(x)+\ln(A)$$=\frac{1}{x}\ln$$Ax\ )=\ln\((Ax)^{\frac{1}{x}}$$$ I checked this equation on the computer, and it gives the solution in the form you gave.
 March 10th, 2011, 02:14 PM #23 Senior Member   Joined: Feb 2011 Posts: 150 Thanks: 0 Re: Exact form, differential equations Ah interesting stuff, useful to know. Thats me done for tonight, thanks again for your help
 March 10th, 2011, 02:17 PM #24 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Exact form, differential equations Anytime!

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