ay (t) + by (t) + cy(t) = 0 (2)

Transcription

1 Solving ay + by + cy = 0 Without Characteristic Equations, Complex Numbers, or Hats John Tolle Department of Mathematical Sciences Carnegie Mellon University Pittsburgh, PA Some calculus courses feature an introuction to ODEs for those stuents who o not nee a full-flege ifferential equations course. First orer linear equations of the form y (t) + p(t)y(t) = q(t) (1) an secon orer linear equations with constant coefficients, of the form ay (t) + by (t) + cy(t) = 0 (2) are typically covere. The stanar approach is to apply a solution metho for (2) which is arguably importe without context from the ifferential equations curriculum. This approach involves the characteristic equation ar 2 + br + c = 0 associate with (2); it requires complex numbers, an it necessarily raises the problematic issue of whether there is a natural way to iscover that y(t) = te rt is a solution to (2) when b 2 ac = 0. However, in the context of a calculus course, one might consier it preferable to bring to bear as much actual ifferentiation theory, an antiifferentiation theory, as possible upon the enterprise of solving ifferential equations. There is a way to solve (2) using only ieas an tools alreay familiar to the calculus stuent, save one piece of unfamiliar theory, namely, that the complete solution set to a linear ODE may be generate by linear combinations of (the right number of) particular solutions. The peagogical approach we avocate here is to solve (1) an then generalize the basic strategy to (2). The metho of solving (2) is not new, an is far from unknown, but it is selom use. It has been relegate to exercises, its peagogical merits largely unrecognize. We reiterate that this approach is best suite to a curriculum in which no ODEs more general than (2) will be consiere, an for which it is preferable to omit any theory which is otherwise only pertinent to ifferential equations courses. We start by noting the similarity of the left han sie of (1) to the right han sie of t (fg) = fg + f g, the Prouct Rule for ifferentiation. If y plays the role of g, then for y + py

2 to be the erivative of an actual prouct, we woul nee f 1, which of course will not work. But if we multiply (1) by an as-yet-unknown an never-zero function w(t), we get wy + wpy = wq, an then we nee w = wp in orer to transform (1) into the equivalent equation (wy) = wq, (3) t which we can then solve via antiifferentiation; in other wors, (3) is a ifferential equation of the simplest type, (unknown function) = given function. t An of course, the multiplier function w which serves as an ingreient to help us reuce (1) to (3) is ( ) w(t) = exp p(t)t. Now let s consier equations of type (2), but let us choose the alternate stanar form y + ay + by = 0. () Now multiply but an as-yet-unknown, never-zero function w(t): wy + awy + bwy = 0. (5) We now have two erivatives involve. Piggybacking on our strategy for (1), we might try reucing (5) to the simple ifferential equation 2 (w(t)y(t)) = 0, t2 which we coul then solve with two antiifferentiation steps. So let s see if that will work. The Prouct Rule for secon erivatives is (fg) = fg + 2f g + f g, so we want to fit the left han sie of (5) into this moul. This woul mean that y shoul play the role of g, an then w shoul play the role of f, just as before. But now we nee both of w = a 2 w w = bw

3 to hol for fg + 2f g + f g = wy + awy + bwy to be true. The first conition requires w(t) = e at/2. But then we woul have w (t) = a2 eat/2 = a2 a2 w(t), an so the secon conition only hols if b =. So it is not generally possible to express the left han sie of (5) as the secon erivative of a prouct. However, we can rewrite (5) as ) wy + awy + (b a2 wy + a2 wy = 0. (6) Then let w(t) = e at/2, an recognize the first three terms of (6), wy + awy + a2 wy, as the secon erivative of w(t)y(t). Therefore (6) becomes 2 ) (b t (w(t)y(t)) + a2 wy = 0. Setting u = wy, we get u + ) (b a2 u = 0. One can point out to the stuents that the preceing steps are similar to the technique of completing the square, use to solve (quaratic) algebraic equations. At this point, one can iscuss how to solve simple secon orer equations of the form u + cu = 0, or the equivalent form u = βu. (7) (The instructor may even wish to iscuss this type of equation much earlier so it is alreay familiar to the stuents.) Regarless of the value of β, the calculus stuent can pretty well solve (7) by inspection. We simply note that if β = 0, solutions are linear an can be obtaine by two antiifferentiations; if β > 0, the exponential solutions u 1 (t) = exp( βt) an u 2 (t) = exp( βt) are easy to fin; an if β < 0, we can easily fin solutions u 1 (t) = cos( βt) an u 2 (t) = sin( βt). So the strategy is to have stuents convert the problem of solving any given instance of () to that of solving an equation of the form (7). Note what happens

4 if β = 0, which correspons to b a2 = 0. Then solutions to (6) are all of the form u(t) = C 1 t + C 2, an then, converting back to y, we get solutions to () all of the form y(t) = C 1 te at/2 + C 2 e at/2. An so we fin the particular solution y(t) = te at/2 without either (i) oing anything special to look for it, e.g., reuction of orer, or using the Wronskian, or (ii) simply proucing it like a rabbit from a hat. All of the calculus texts in the references o (i) or (ii). If β 0, then solving (7) is not quite as simple as performing two antiifferentiations. But we are able to fin two particular solutions, whether of exponential or trigonometric type. So, just as in the usual treatment of (2) in calculus courses, we o nee one piece of ifferential equations theory at this point, namely the result that all solutions to a secon orer linear ODE may be generate by linear combinations of two particular solutions, neither a constant multiple of the other. The only memorization stuents must o to apply this solution approach is to remember to multiply () through by e at/2. An complex numbers are not neee; in particular, we never nee to explain to stuents why a function like y(t) = e (α+iβ)t shoul make any sense to anyone. Ultimately we seek real-value solutions, so not having to eal with complex-value solutions, an complex linear combinations thereof, is a plus. Certainly the characteristic equation approach, generalizable as it is to higher orer ODEs, together with the complex-value solutions one must amit, is useful an powerful in a ifferential equations course. But the approach is cumbersome an unnatural if () is the terminus of one s ODE stuies. Perhaps some calculus instructors will fin the metho outline here to be an attractive alternative. Acknowlegments Thanks to John Mackey an Tim Flaherty for their interest, an for useful iscussion. The author learne the above approach from Exercise of []. References 1. D.D. Berkey an P. Blanchar, Calculus, 3r e., Sauners College Publishing, C.H. Ewars, an D.E. Penney, Calculus, 6th e., Prentice Hall, D. Hughes-Hallet, A.M. Gleason, et al., Calculus, John Wiley & Sons, 199.

5 . R.A. Moore, Introuction to Differential Equations, Allyn an Bacon, J. Stewart, Calculus: Early Transcenentals, th e., Brooks-Cole, M.J. Strauss, G.L. Braley, an K.J. Smith, Calculus, 3r e., Prentice Hall, 2002.