Complex roots of the characteristic equation

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Complex roots of the characteristic equation

  1. 1. Complex Roots of the Characteristic EquationWe established previously that if we had a solution of the form ert to the second orderequation ay ′′ + by ′ + cy = 0then r must satisfy ar 2 + br + c = 0which we called the characteristic equation. We also established that for a second-degree linear homogeneous equation, we needed 2 linearly independent solutions towrite down the general solution, and that a non-zero Wronskian would tell us whenwe had linearly independent solutions.In the case where the characteristic equation has two real roots (r1 and r2 ), we gettwo linearly independent solutions (er1 t and er2 t ) and therefore the general solution.In the case with one repeated real root r, we used reduction of order to find twolinearly independent solutions: ert and tert .Today we discuss how to deal with complex roots of the characteristic equation: 1. An Unsatisfactory Result 2. Pure Imaginary Roots 3. Complex Roots 4. Amplitude and Period 5. Practice1 An Unsatisfactory ResultLet us attempt to find solutions to the equation y ′′ + y = 0, using our usual methodof looking for solutions of the form ert .Here, the characteristic equation is r 2 + 1 = 0, which has solutions r = ±i, where i is √the imaginary number i = −1. Hence, we could get two “solutions” y1 = eit , y2 = e−it 1
  2. 2. and a “general” solution y(t) = c1 eit + c2 e−it .However, this is extremely unsastifactory. We started with a real differential equation,and ended with a complex function as a solution! We want to have a real solution.It is worth noting that we can use Euler’s identity eit = cos(t) + i sin(t)to rewrite these solutions into terms of sines and cosines, and thus make the imaginarypart of the solution explicit.We will now attempt to find some real valued solutions to differential equations withcomplex roots to the characteristic equation.2 Pure Imaginary RootsFor the equation y ′′ + y = 0 we have found two possible “solutions”: y1 = eit , and y2 = e−it .or y1 (t) = cos(t) + i sin(t) y2 (t) = cos(−t) + i sin(−t) = cos(t) − i sin(t)Now let us apply the principal of superposition: given that these two functions aresolutions, so is any linear combination of the two. Particularly, we have one solutionof the form 1 it 1 e + e−it = [(cos(t) + i sin(t)) + (cos(t) − i sin(t))] 2 2 1 = [2 cos(t)] 2 = cos(t)Eureka! The function cos(t) is a real function, and is a solution to y ′′ + y = 0! (Infact, it’s easy enough to check this directly.)To get our second solution, we need to multiply both equations by i/2 and thensubtract in the right order: i −it i i2 i i2 (e − eit ) = cos(t) − sin(t) − cos(t) − sin(t) 2 2 2 2 2 = −i2 sin(t) = sin(t)So (perhaps not surprisingly) the other solution is sin(t). (Note: The approach wehave taken here will work any time we have a solution and its complex conjugate.
  3. 3. In other words: If y = a(t) + ib(t) and y = a(t) − ib(t) are both solutions to ahomogeneous equation, the principle of superposition guarantees that both a(t) andb(t) are solutions to the differential equation also.)Are these two solutions linearly independent? We check the Wronskian: cos(t) sin(t) = cos2 (t) + sin2 (t) = 1 − sin(t) cos(t)So these are linearly independent, and the general solution to y ′′ + y = 0 must be y(t) = c1 cos(t) + c2 sin(t).Example:Find two real fundamental solutions to y ′′ + 3y = 0, and show that they are linearlyindependent.The characteristic equation is r 2 + 3r = 0 √ √ √ √which has roots ± 3i. Thus we get two solutions, e± 3t i = cos( 3 t) ± i sin( 3 t).The real and imaginary parts give two real solutions √ √ y1 (t) = cos( 3 t) and y2 = sin( 3 t).We can see that these are linearly independent as the Wronskian is √ √ cos( 3 t) sin( √t) 3 √ √ √ √ √ √ √ √ = 3 cos2 ( 3 t) + 3 sin2 ( 3 t) = 3 − 3 sin( 3 t) 3 cos( 3 t)which is not zero.It is worth noting that complex roots to polynomials always come in conjugate pairs:if a + bi is a solution, then so is a − bi. (This is a fact that you encountered brieflyin Vector Geometry.) So if we have two purely imaginary roots, they must be of theform ±bi, with b real and not zero.Then our real and imaginary parts of e±bt i are cos(b t) and sin(b t), and the Wronskianfor these is non-zero: cos(b t) sin(b t) = b cos2 (b t) + b sin2 (b t) = b −b sin(b t) b cos(b t)Thus, we have completely solved the case in which we have purely imaginary roots.There will always be two linearly independent real solutions of the form cos(bt) andsin(bt) which we can use to write down the general solution.
  4. 4. 3 Complex RootsLet us now suppose that our characteristic equation has a complex root. In otherwords, we will have a root of the form a + bi where a and b are real and non-zero. Ofcourse, since complex roots come in conjugate pairs, a − bi is also a root. Thus, wehave the following two solutions to the original differential equation: e(a+bi)t = eat ebt i = eat cos(bt) + ieat sin(bt) and e(a−bi)t = eat e−bt i = eat cos(bt) − ieat sin(bt)Using exactly the same trick as in the pure imaginary case, we extract the real andimaginary parts of these solutions: 1 (a+bi)t e + e(a−bi)t = eat cos(bt) and 2 i (a−bi)t e − e(a+bi)t = eat sin(bt) 2Will these two solutions be linearly independent? We check the Wronskian: eat cos(bt) eat sin(bt) eat (a cos(bt) − b sin(bt)) eat (b cos(bt) + a sin(bt)) = e2at (b cos2 (bt) + a cos(bt) sin(bt)) − e2at (a sin(bt) cos(bt) − b sin2 (bt)) = be2atwhich is nonzero, since we are assuming b = 0.Example:Solve the initial value problem y ′′ − 2y ′ + 2y = 0, y(π) = eπ , y ′ (π) = 0.The characteristic equation iswhich has rootsSo our fundamental solutions are y1 (t) =and y2 (t) =for a general solution
  5. 5. y(t) =Then we use the initial conditions. y(π) = eπ means thatThen y ′(π) = 0 means thatSo the unique solution to this initial value problem is4 Amplitude and PeriodSuppose we have a differential equation which has a solution of the form y(t) = c1 cos(bt) + c2 sin(bt)Then we have a periodic solution, and we sometimes want to know what the amplitudeand period of our solution is. We will answer this by rewriting our solution above inthe form y = R cos(ωt − δ)Then our amplitude will be R, the period will be 2π/ω, and δ will be a phase shift.(It simply shifts the graph right or left of a typical cosine. The graph of cos(ωt − δ)is shifted right of the graph of cos(ωt) by the amount δ/ω.)We can rewrite the solution using the fact that cos(A − B) = cos(A) cos(B) + sin(A) sin(B)In our case, we want R cos(ωt − δ) = R cos(δ) cos(ωt) + R sin(δ) sin(ωt) = c1 cos(bt) + c2 sin(bt)In other words, we want ω = b, R cos(δ) = c1 , and R sin(δ) = c2 . So we know ω andwe get R 2 = c2 + c2 1 2 c2 tan(δ) = c1
  6. 6. If all we want is the amplitude and period, we do not need to actually solve for thephase shift δ. (Note that solving for R and δ like this is exactly like finding the radiusand angle to put the point (c1 , c2 ) into polar coordinates. Caution: You will need tocheck the signs of c1 and c2 to make sure to choose δ in the correct quadrant.)Example:Solve the initial value problem y ′′ + 2y = 0, y(0) = 1, and y ′ (0) = 3. Then find theamplitude and period of the solution. √The characteristic equation is r 2 + 2 = 0, so r = ± 2i. Thus the general solution is √ √ y(t) = c1 cos( 2t) + c2 sin( 2t)The initial condition y(0) = 1 gives us c1 = 1, and since √ √ √ √ y ′(t) = − 2 sin( 2t) + c2 2 cos( 2t) √we see that y ′(0) = 3 means that c2 = 3/ 2, so we have the solution √ 3 √ y(t) = cos( 2t) + √ sin( 2t). 2To find the amplitude, we simply need 2 3 9 11 R= 12 + √ = 1+ = ≈ 2.345. 2 2 2 √ √The period is 2π/ω, where here ω = 2, so the period is 2 π.We can also write down the frequency, which is the reciprocal of the period: 1 1 frequency = =√ period 2πFinally, we sometimes refer to the angular frequency or radian frequency, which is thefrequency multiplied by 2π. (In other words, how often we repeat within a circle of2π radians. If you follow all the twists and turns here, angular frequency is just ω.) √Here of course the angular frequency is ω = 2.We can rewrite our solution in the form (approximately) √ y(t) = 2.345 cos( 2t − δ) 3where we can determine that since tan(δ) = √2 and we have (c1 , c2 ) in the first √quadrant, δ = tan−1 (3/ 2) ≈ 1.13. (If c1 and c2 had both been negative, we would √have required a δ in the third quadrant, so we would have had tan−1 (3/ 2) + π ≈4.27.)
  7. 7. 5 PracticeNow find general solutions to each of the following: • y ′′ + 6y ′ + 13y = 0 • y ′′ + 2y ′ − 3y = 0 • 4y ′′ − 4y ′ + y = 0

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