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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 
Newbie Joined: Jun 2015 From: House Posts: 11 Thanks: 0  
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 $${\mathrm d^2 y \over \mathrm d x^2} + p{\mathrm d y \over \mathrm d x} + q = 0$$ 
June 11th, 2015, 06:21 AM  #5  
Newbie Joined: Jun 2015 From: House Posts: 11 Thanks: 0  Quote:
(u1)(u2) (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. $$\big(u  (1+2\mathrm i)\big)\big(u  (12\mathrm i)\big)$$ 
June 11th, 2015, 07:24 AM  #7 
Newbie Joined: Jun 2015 From: House Posts: 11 Thanks: 0  
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. 

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