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 October 11th, 2010, 06:37 PM #1 Member   Joined: Sep 2010 Posts: 73 Thanks: 0 you are given general solution - find differential equation You are given the general solution - find the homogeneous 2nd order differential equation ay'' + by' + cy = 0. y(x) = c_1 + c_2 * e^(-10x) So I got y'(x) = -10 c_2 e^(-10x) and y''(x) = 100 c_2 e^(-10x) And I see from the given general solution that e^(0x) = 1 and e^(-10x) are two particular solutions, y_1 and y_2. So given that the book says we are supposed to guess y_1 = e^(r_1 * x) and y_2 = e^(r_2 * x) [I still don't understand WHY we have to 'guess' this], that means my r_1 = 0 and my r_2 = -10. Now.. how do I find the differential equation from here?
 October 11th, 2010, 07:27 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 y'' + 10y' = 0
 October 11th, 2010, 07:47 PM #3 Member   Joined: Sep 2010 Posts: 73 Thanks: 0 Re: you are given general solution - find differential equat Yes, I know that's the answer, but HOW do I get there from where I am?
 October 11th, 2010, 08:18 PM #4 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: you are given general solution - find differential equat You are told your auxiliary equation has roots 0 and -10, thus r² + 10r = 0 Since e^(rx) cannot be zero, multiply through. (r² + 10r)e^(rx) = 0 (a) r²e^(rx) + 10re^(rx) = 0 We are given y = c1 + c2e^(rx), thus y' = c2re^(rx) y'' = c2r²e^(rx) Thus, substituting into (a), we may state: y''/c2 + 10y'/c2 = 0 Multiply through by c2. y'' + 10y' = 0
 October 12th, 2010, 02:10 AM #5 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 You can go directly from the auxiliary equation to your answer (or vice versa). Up to multiplication by a constant (and rearrangement of terms), the result is unique. [color=#00AA00]MarkFL[/color] was aiming for a rigorous proof, but overlooked the possibility that c2 = 0.
 October 12th, 2010, 09:20 AM #6 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: you are given general solution - find differential equat Yes, I did make the assumption that c2 ? 0 thereby missing the trivial solution y(x) = c1. I was just trying to show a path to the non-trivial solution given by [color=#00BF00]skipjack[/color].
 October 12th, 2010, 10:29 AM #7 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: you are given general solution - find differential equat Lol, I really should get my brain in gear before typing away. I ignored, rather than missed the trivial solution. And the equation given encompasses all solutions, trivial or otherwise. In other words, ignore what I said.
 October 12th, 2010, 01:09 PM #8 Global Moderator   Joined: Dec 2006 Posts: 20,931 Thanks: 2205 Multiplying r² + 10r = 0 by c2e^(rx) gives c2e^(rx)r² + 10c2e^(rx)r = 0 (whether or not c2e^(rx) = 0). Now making the substitutions you suggested gives y'' + 10y' = 0.

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