Complex Solutions to Real Solutions and a Computation

Complex Solutions to Real Solutions and a Computation 1 Sines and Cosines We begin with a “side computation”. Using Euler’s Formula, eibt = cos(bt) + i sin(bt), we see that: eibt + e−ibt = 2 = = cos(bt) + i sin(bt) + cos(bt) − i sin(bt) 2 2 cos(bt) 2 cos(bt) Similarly, eibt − e−ibt = 2i = = 2 cos(bt) + i sin(bt) − cos(bt) + i sin(bt) 2i 2i sin(bt) 2i sin(bt) The Solution Now we consider solutions to the second order, linear, homogeneous differential equation: ay 00 + by 0 + cy = 0 From the characteristic equation, ar2 +br+c = 0, we suppose that the solutions are complex conjugates: r = s ± iw Note that in this case, while s could be zero, part of the assumption is that w 6= 0 (Otherwise, we would have one real solution to the characteristic equation). Then, as usual, we take y = c1 e(s+iw)t + c2 e(s−iw)t . We show that a fundamental set of solutions is given by: y1 (t) = est cos(wt), y2 (t) = est sin(wt) which removes the complex number i (in fact, any solution to an initial value problem will formulate the c1 , c2 in such a way as to remove the complex number i). From the general form of the solution, solve the specific initial value problems, with initial values: y1 (0) = 1 y2 (0) = 0 y10 (0) = s y20 (0) = w Then, we see that the Wronskian will be non-zero for any valid values of s, w. Let’s look at y1 first: If y1 (0) = 1, then c1 + c2 = 1. The second condition gives: y10 (0) = s s(c1 + c2 ) + ib(c1 − c2 ) = s ib(c1 − c2 ) = 0 c1 − c2 = 0 so : (c1 + c2 = 1) From which we get that c1 = c2 = 21 . Then: y1 (t) = e(s+iw)t + e(s−iw)t est (eiwt + e−iwt ) = = est cos(wt) 2 2 1 We now consider y2 : If y2 (0) = 0, then c1 + c2 = 0. The second condition gives: y20 (0) = w s(c1 + c2 ) + iw(c1 − c2 ) = w iw(c1 − c2 ) = w 1 c1 − c2 = i From which we get that c1 = y2 (t) = 1 2i , c2 so : (c1 + c2 = 0) 1 = − 2i . Then: est (eiwt − e−iwt ) e(s+iw)t − e(s−iw)t = = est sin(wt) 2i 2i We now conclude with the following. If we are given: ay 00 + by 0 + cy = 0 with roots to the characteristic equation: ar2 + br + c = 0 complex conjugates, r = s ± iw, then the general solution to the seond order linear homogeneous differential equation is: yh (t) = est (c1 cos(wt) + c2 sin(wt)) (Note the absence of the complex number i). 3 Analysis of the Solution It is convenient to rewrite: c1 cos(wt) + c2 sin(wt) as a single periodic function. This will allow for a quick sketch of the solution, and the analysis will be easier than for the sum. Using the trig identity: R cos(ωt − δ) = R cos(δ) cos(ωt) + R sin(δ) sin(ωt) we see that c1 = R cos(δ), c2 = R sin(δ). From this, we get that, to convert: c1 cos(wt) + c2 sin(wt) = R cos(ωt − δ) p we set R = c21 + c22 and δ = Tan−1 (c2 /c1 ). Note that the inverse tangent is: • Taken on the sine coefficient divided by the cosine coefficient. • δ is the actual angle- that is, from the signs of c1 and c2 , we need to know which quadrant δ is in, and make the appropriate change. Re-writing the solution then yields: y(t) = Rest cos(ωt − δ) The pseudoperiod of y is 2π ω , and the phase angle is δ. 2