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 November 8th, 2016, 09:04 AM #1 Newbie   Joined: Nov 2016 From: London Posts: 3 Thanks: 0 First order differential equation Hey all, I have a separable differential equation: Can someone please explain how the solution is: Thanks
 November 8th, 2016, 09:25 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra Post your attempt first. There's nothing that should cause a differential equations student much trouble, especially if you know how to integrate by partial fractions (writing your constant of integration as $\ln c$) and how to use logarithmic identities to good effect. Perhaps the hardest part is removing the absolute value signs.
 November 8th, 2016, 11:09 AM #3 Newbie   Joined: Nov 2016 From: London Posts: 3 Thanks: 0 I get 1/2ln(2x+1)=ln(t/(t+1))+ln(C), but what should I do then? Last edited by skipjack; November 8th, 2016 at 02:49 PM. Reason: to insert parentheses
 November 8th, 2016, 02:54 PM #4 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 Can you write that equation in the form ln(...) = ln(...)?
 November 8th, 2016, 03:38 PM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra You are looking for the terms in $t$ to be squared, so multiply by 2 and use properties of the logarithm. Last edited by skipjack; November 8th, 2016 at 03:41 PM.
 November 8th, 2016, 03:42 PM #6 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 Mrkukas, are you given that t > 0?
 November 8th, 2016, 04:42 PM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,313 Thanks: 2447 Math Focus: Mainly analysis and algebra He shouldn't be. It should be a valid result for all $t$ with the possible exception of $t=-1$.
 November 8th, 2016, 06:16 PM #8 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 That depends on what you mean by "valid result" and "possible exception". The original equation doesn't imply that t isn't -1, but the supplied solution does, as it's undefined at t = -1.
November 8th, 2016, 11:34 PM   #9
Newbie

Joined: Nov 2016
From: London

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Quote:
 Originally Posted by skipjack Mrkukas, are you given that t > 0?
No, I am not.

Last edited by skipjack; November 8th, 2016 at 11:51 PM.

 November 8th, 2016, 11:53 PM #10 Global Moderator   Joined: Dec 2006 Posts: 19,059 Thanks: 1619 Can you assume that t is real?

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