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June 11th, 2015, 04:19 AM   #1
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Writing equation with complex roots

I was given two complex numbers as roots of differential equation:
k1, k2 = 1 +- 2i

I need to write a equation.

Am I right, that I need to write
y = e^x(C1cos2x+C2sin2x) then get two derivatives and put them in a format
y''+py'+qy = 0

?
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June 11th, 2015, 04:42 AM   #2
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Do you mean that the complex numbers are roots of the characteristic equation? If so, then your $$y = \mathrm e^x(C1\cos2x+C2\sin2x)$$is the general solution of the homogeneous equation.
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June 11th, 2015, 04:46 AM   #3
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Quote:
Originally Posted by v8archie View Post
Do you mean that the complex numbers are roots of the characteristic equation? If so, then your $$y = \mathrm e^x(C1\cos2x+C2\sin2x)$$is the general solution of the homogeneous equation.
Yes, but I actually dont need solution, but to get equation from these.
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June 11th, 2015, 05:26 AM   #4
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Your approach would yield the correct answer.

But you can write the equation from knowledge about the characteristic equation without knowing the solution.

If the roots of the characteristic equation are $\alpha$ and $\beta$ then the characteristic equation is $(u - \alpha)(u - \beta)$. This can be expanded into the form $u^2 + pu + q$ and then the associated homogeneous differential equation would be $${\mathrm d^2 y \over \mathrm d x^2} + p{\mathrm d y \over \mathrm d x} + q = 0$$
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June 11th, 2015, 06:21 AM   #5
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Quote:
Originally Posted by v8archie View Post
Your approach would yield the correct answer.

But you can write the equation from knowledge about the characteristic equation without knowing the solution.

If the roots of the characteristic equation are $\alpha$ and $\beta$ then the characteristic equation is $(u - \alpha)(u - \beta)$. This can be expanded into the form $u^2 + pu + q$ and then the associated homogeneous differential equation would be $${\mathrm d^2 y \over \mathrm d x^2} + p{\mathrm d y \over \mathrm d x} + q = 0$$
So in my case the characteristic equation is:
(u-1)(u-2)

(u^2)-3u+2 = 0

y''-3y'+2 = 0 ?
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June 11th, 2015, 06:31 AM   #6
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No, your roots are complex, not real.
$$\big(u - (1+2\mathrm i)\big)\big(u - (1-2\mathrm i)\big)$$
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June 11th, 2015, 07:24 AM   #7
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Quote:
Originally Posted by v8archie View Post
No, your roots are complex, not real.
$$\big(u - (1+2\mathrm i)\big)\big(u - (1-2\mathrm i)\big)$$
So it would be:
u^2-2u+5 = 0

u''-2u'+5=0

?

i think
(1-2i)+(1+2i) = 2+0i=2
(1-2i)*(1+2i) = 5+0i = 5
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June 11th, 2015, 08:33 AM   #8
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Correct, though I'd use y for the independent variable of the homogeneous differential equation:
y'' - 3y' + 2 = 0.
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