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 Differential Equations Ordinary and Partial Differential Equations Math Forum

 June 11th, 2015, 04:19 AM #1 Newbie   Joined: Jun 2015 From: House Posts: 11 Thanks: 0 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 ? June 11th, 2015, 04:42 AM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 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. June 11th, 2015, 04:46 AM   #3
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 Originally Posted by v8archie 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. June 11th, 2015, 05:26 AM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 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 $${\mathrm d^2 y \over \mathrm d x^2} + p{\mathrm d y \over \mathrm d x} + q = 0$$ Thanks from dudeperfect June 11th, 2015, 06:21 AM   #5
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 Originally Posted by v8archie 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 ? June 11th, 2015, 06:31 AM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra No, your roots are complex, not real. $$\big(u - (1+2\mathrm i)\big)\big(u - (1-2\mathrm i)\big)$$ June 11th, 2015, 07:24 AM   #7
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 Originally Posted by v8archie 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 June 11th, 2015, 08:33 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2203 Correct, though I'd use y for the independent variable of the homogeneous differential equation: y'' - 3y' + 2 = 0. Thanks from dudeperfect Tags complex, equation, roots, writing Search tags for this page
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# e^x (c1cos2x c2sin2x)

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