Complex Roots relate to the topic of Second order Linear Homogeneous equations with constant coefficients. The Second Order linear refers to the equation having the setup formula of y”+p(t)y’ + q(t)y = g(t). Constant coefficients are the values in front of the derivatives of y and y itself. Homogeneous means the equation is equal to zero.So a homogeneous equation would look like
y”+by’ + cy = 0 or y”+p(t)y’ + q(t)y = 0.
When solving the general solution of the homogeneous equation, we have to look for the characteristic equation which is r^2 + br + c = 0. But how would you get there? To start we are given the differential equation of
y = e^rt
which can be used to look for the characteristic equation. To start we look for the derivative of the differential equation then replace y and its derivatives with what we found which the homogeneous equation would look like
(r^2)e^rt + bre^rt + ce^rt = 0
You can see there is a common value which is e^rt you can factor out. Once you factor it out, you can divide e^rt and this is the characteristic equation,
r^2 + br + c = 0
The general solution of the homogeneous equation will have 3 different results depending on the roots which are found from the characteristic equation. This is where Complex Roots come into play. Complex Roots is when the roots have an imaginary number or i which looks like r = a ± i*b. The general solution will be
y = C1e^(a + ib)t + C2e^(a – ib)t
But there is a problem when you want to solve for a particular solution how would you get rid of the i? So the general solution is still not done. With the general solution, y = C1e^(a + ib)t + C2e^(a – ib)t you can right away separate the roots which will become y = C1e^at * e^(ibt) + C2e^at * e^( –ibt). Next you factor out e^a which makes the equation y = e^at(C1e^(ibt) + C2e^(-ibt). This is where the Euler’s Formula is introduced which says that
e^it = cos(t) + isin(t)
So the equation is written as such
y = e^at(C1(cos(bt) + isin(bt)) + C2(cos(-bt) + isin(-bt)))
This is where your knowledge of trig identities comes to play. Two trig identities which can be used are
sin(-x) = -sin(x), cos(-x) = -cos(x)
This would replace the negative values in the sin and cos but we are left with the complex part i. To get rid of the complex part, the equation is rearranged so that the constants can be created into a new constant that will work with real and imaginary numbers.
y = e^at((C1 + C2)cos(bt) + (C1i – C2i)isin(bt)))
y = e^at(Acos(bt) + Bsin(bt)), general solution
Sample Problem: y” + 4y’ + 5y = 0, y(0) = 1, y'(0) = 0 y=e^(rt), y’ = re^(rt), y” = (r^2)e^(rt) e^(rt)*(r^2 + 4r +5) = 0
Since the roots are the easy to separate, we use the quadratic formula to look for the roots which you will end up with r = -2±i
The General Solution will equal to
y = e^(-2t)*(Acos(t) + Bsin(t))
For the particular solution, we solve for A and B using y(0) = 1 and y'(0) = 0. y(0) = 1 1 = e^(-2(0))*(Acos(0) + Bsin(0)) 1 = A
y'(0) = 0 y’ = -2Ae^(-2t)cos(t) – Ae^(-2t)sin(t) – 2Be^(-2t)sin(t) + Be^(-2t)cos(t) 0 = -2(1)e^(-2(0))cos(0) – (1)e^(-2(0))sin(0) – 2Be^(-2(0))sin(0) + Be^(-2(0))cos(0) 0 = -2 + B B = 2
The Particular Solution:
y = e^(-2t)*(cos(t) + 2sin(t))
Videos to help with Studies:
Complex Roots Part 1
Complex Roots Part 2
Complex Roots Part 3
