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 August 29th, 2010, 11:35 AM #1 Newbie   Joined: Aug 2010 Posts: 13 Thanks: 0 Help working out this differential equation... Hi, I'm currently working on some differential equations. I'm stuck on one and there's a few more equations similar to it. I was wondering, can someone show me a step by step of working out this equation out... x'' (t) - 4 x' (t) + 3 x (t) = 0 If I can get a step by step of the of it, i'll be able to work out the other equations similar to it. Thank you.
August 29th, 2010, 12:54 PM   #2
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Re: Help working out this differential equation...

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 Originally Posted by Britmouth Hi, I'm currently working on some differential equations. I'm stuck on one and there's a few more equations similar to it. I was wondering, can someone show me a step by step of working out this equation out... x'' (t) - 4 x' (t) + 3 x (t) = 0 If I can get a step by step of the of it, i'll be able to work out the other equations similar to it. Thank you.
The usual approach is to assume x(t)=exp(at). The the differential equation becomes an algebraic (quadratic) in a.

a² -4a + 3=0. So a=3 and a=1 give the two solutions.

August 29th, 2010, 01:12 PM   #3
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Re: Help working out this differential equation...

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 Originally Posted by mathman The usual approach is to assume x(t)=exp(at). The the differential equation becomes an algebraic (quadratic) in a. a² -4a + 3=0. So a=3 and a=1 give the two solutions.
Thanks but I still don't understand.

Is it possible you can be a bit more detailed?

 August 29th, 2010, 02:10 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: Help working out this differential equation... What mathman is saying is that if we assume x(t) = k?e^(at), then we have x'(t) = a?k?e^(at) = a?x(t) and x''(t) = a²?k?e^(at) = a²?x(t) Substituting these into the original ODE, we get a²?x(t) - 4(a?x(t)) + 3(x(t)) = 0 Factoring gives x(t)?(a² - 4a + 3) = x(t)(a - 1)(a - 3) = 0 The expression a² - 4a + 3 is referred to as the auxiliary equation. Notice that x(t) = 0 is a solution, but it is referred to as a trivial solution. Thus, to get two independent non-trivial solutions to the ODE, we can use the roots of the auxiliary equation to get f(t) = k1?e^(t) and g(t) = k2?e^(3t) Thus we have f''(t) - 4?f'(t) + 3?f(t) = k1?e^(t) - 4?k1?e^(t) + 3?k1?e^(t) = 0 and g''(t) - 4?g'(t) + 3?g(t) = 9?k2?e^(3t) - 4(3?k2?e^(3t)) + 3(k2?e^(3t)) = 0 Adding these two equations together, we have [f''(t) + g''(t)] - 4[f'(t) + g'(t)] + 3[f(t) + g(t)] = 0 If we let h(t) = f(t) + g(t), and using the fact that the nth derivative of h is the sum of the nth derivatives of f and g, we may state h'(t) = f'(t) + g'(t) h''(t) = f''(t) + g''(t) Thus, we have h''(t) - 4?h'(t) + 3?h(t) = 0 where h(t) is considered the general solution (can be shown using the existence and uniqueness theorem), and is given by h(t) = k1?e^(t) + k2?e^(3t) Note: if the roots of the auxiliary equation are complex, then the solution will be sinusoidal.
 August 29th, 2010, 02:53 PM #5 Newbie   Joined: Aug 2010 Posts: 13 Thanks: 0 Re: Help working out this differential equation... Thanks a lot MarkFL! That's what I was looking for. I can use your example to work out the other equations.
 August 29th, 2010, 09:07 PM #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: Help working out this differential equation... I was looking over my old DiffEq textbook and noticed I failed to mention that if the discriminant of the auxiliary equation is zero, giving a repeated root r, then the method of reduction of order must be used to yield the general solution x(t) = k1·e^(rt) + k2·t·e^(rt) Sorry for that omission.
 August 30th, 2010, 12:01 AM #7 Global Moderator   Joined: Dec 2006 Posts: 20,805 Thanks: 2150 Solving the quadratic equation mentioned by mathman allows an integrating factor to be found: x'' - 4x' + 3x = 0 e^(-t)x'' - 4e^(-t)x' + 3e^(-t)x = 0 d(e^(-t)x' - 3e^(-t)x)/dt = 0 e^(-t)x' - 3e^(-t)x) = A, where A is a constant e^(-3t)x' - 3e^(-3t)x) = Ae^(-2t) d(e^(-3t)x)/dt = Ae^(-2t) e^(-3t)x = Ae^(-2t) + B, where B is a constant x = Ae^t + Be^(3t)
 August 30th, 2010, 05:40 AM #8 Newbie   Joined: Aug 2010 Posts: 13 Thanks: 0 Re: Help working out this differential equation... Thanks skipjack and MarkFL! I've just attempted some differentiation and integration equations. I was wondering if you can check it for me, to see whether i'm doing it right. differentiate: $x^3 + 2x^3 - 3$ my answer: $3 + 6x^2$ $e^3^x$ my answer: $3e^3^x$ $sin 2x - 2 cos 3x$ my answer: $2 cos 2x - (-)6 sin 3x$ integrate: $x^2 + 3x - 2$ my answer: = $x^3/3 + 3x^2/2 - 2x$ $e^2^x$ my answer: $e^2^x/2$ $cos(x/2)$ my answer: $2 sin (x/2)$ If I've done something wrong, can you explain where I went wrong.
 August 30th, 2010, 06:14 AM #9 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: Help working out this differential equation... The only one I see a problem with is the first one: d(x^3 + 2x^3 - 3)/dx = d(3x^3 - 3)/dx = 9x^2 Given that this was the easiest one, and you did the others correctly, did you copy the problem correctly? Looks like you should have began with 3x + 2x^3 - 3 instead... For the integrations, you might want to add a constant to the results. skipjack: as always, very clever!
August 30th, 2010, 07:57 AM   #10
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Re: Help working out this differential equation...

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 Originally Posted by MarkFL The only one I see a problem with is the first one: d(x^3 + 2x^3 - 3)/dx = d(3x^3 - 3)/dx = 9x^2 Given that this was the easiest one, and you did the others correctly, did you copy the problem correctly?
No the equation was written properly.

I guess I should of added 3 and 6 together. Thanks for the correction.

BTW, what do you mean by add contants to the integration?

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