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August 29th, 2010, 11:35 AM   #1
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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.
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August 29th, 2010, 12:54 PM   #2
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Re: Help working out this differential equation...

Quote:
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.

a -4a + 3=0. So a=3 and a=1 give the two solutions.
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August 29th, 2010, 01:12 PM   #3
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Re: Help working out this differential equation...

Quote:
Originally Posted by mathman

The usual approach is to assume x(t)=exp(at). The the differential equation becomes an algebraic (quadratic) in a.

a -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?
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August 29th, 2010, 02:10 PM   #4
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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) = a?k?e^(at) = a?x(t)

Substituting these into the original ODE, we get

a?x(t) - 4(a?x(t)) + 3(x(t)) = 0

Factoring gives

x(t)?(a - 4a + 3) = x(t)(a - 1)(a - 3) = 0

The expression a - 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.
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August 29th, 2010, 02:53 PM   #5
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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.
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August 29th, 2010, 09:07 PM   #6
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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) = k1e^(rt) + k2te^(rt)

Sorry for that omission.
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August 30th, 2010, 12:01 AM   #7
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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)
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August 30th, 2010, 05:40 AM   #8
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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:
my answer:
my answer:
my answer:

integrate:
my answer: =
my answer:
my answer:

If I've done something wrong, can you explain where I went wrong.
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August 30th, 2010, 06:14 AM   #9
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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!
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August 30th, 2010, 07:57 AM   #10
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Re: Help working out this differential equation...

Quote:
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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