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4th order DE

by zoom1
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zoom1
#1
Mar3-12, 05:51 PM
P: 23
I have a 4th order differential equation with given -2 +3i root.

Now need to find the homogenous solution. Well, if the root was real, it would be easier but now I'm stuck and don't know how to proceed.

What am I supposed to do to solve this ?

Equation is : d4y(t)/dt4 +6d3y(t)/dt3 + 22d2y(t)/dt2 + 30dy(t)/dt + 13y(t) = f(t)

Just need to solve the homogenous part so f(t) is just a dummy function
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AlephZero
#2
Mar3-12, 06:45 PM
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The coefficients are real, so actually you have been given two complex roots, not one.

If you haven't done any formal courses about roots of polynomials, think about what you get when you solve a quadratic equation with a pair of complex roots. That should lead you to finding a quadratic factor (with real coefficients) of the 4th-order equation.
zoom1
#3
Mar4-12, 04:42 AM
P: 23
Quote Quote by AlephZero View Post
The coefficients are real, so actually you have been given two complex roots, not one.

If you haven't done any formal courses about roots of polynomials, think about what you get when you solve a quadratic equation with a pair of complex roots. That should lead you to finding a quadratic factor (with real coefficients) of the 4th-order equation.
You're right there are two roots are given -2 - 3i and -2 + 3i
I tried to recreate a quadratic equation with those roots by assuming coefficient of a=1, b=4 and c=13. However it didn't work.

zoom1
#4
Mar4-12, 04:54 AM
P: 23
4th order DE

Ok, I've just done it, thanks though.
HallsofIvy
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Mar5-12, 01:09 PM
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Great! For those who are interested, though, let me note that since [itex]-2+3i[/itex] and [itex]-2-3i[/itex] are roots then [itex](x- (-2+3i))(x-(-2-3i))= (x+2- 3i)(x+2+ 3i)= (x+2)^2- (3i)^2= x^2+ 4x+ 4+ 9= x^2+ 4x+ 13[/itex]. Now divide [itex]x^4+ 6x^3+ 22x^2+ 30x+ 13[/itex] by that to find the quadratic equation the other roots must satisfy.


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