Higher Order Equations

Transcription

1 Higher Order Equations We briefly consider how what we have done with order two equations generalizes to higher order linear equations. Fortunately, the generalization is very straightforward: 1. Theory. Finding Roots 3. Characteristic Equations and Fundamental Solutions 1 Theory The existence theorem for order n linear differential equations contains no suprises: Theorem 3.5: (p. 188) Let p 0 (t), p 1 (t),...,p n 1 (t) and g(t) be continuous functions defined on a < t < b, and let t 0 be in (a, b). Then the initial value problem y (n) + p n 1 (t)y (n 1) + + p (t)y + p 1 (t)y + p 0 (t)y = g(t) y(t 0 ) = y 0, y (t 0 ) = y 0, y (t 0 ) = y 0,...,y (n 1) (t 0 ) = y (n 1) 0 has a unique solution defined on the entire interval (a, b). So suppose we have an n th order linear homogeneous differential equation y (n) (t) + p 1 (t)y (n 1) (t) + p (t)y (n ) (t) + + p n (t)y(t) = 0 defined on some interval I where the p i are continuous and we are therefore guaranteed a solution. Then we will need n fundamental solutions y 1,...,y n, which are to be linearly independent. In that case, we have a general solution y(t) = c 1 y 1 (t) + c y (t) + c n y n (t). 1

2 To check that our proposed fundamental solutions y i are in fact linearly independent, we use the Wronskian, which now takes the form W(y 1, y,...y n )(t) = y 1 (t) y (t) y n (t) y 1 (t) y (t) y n (t) y (n 1) 1 (t) y (n 1) (t) y n (n 1) (t) (To help you remember this, just notice that what you are doing is setting up a matrix equation to solve for the initial conditions y(t 0 ) = y 0 through y (n 1) (t 0 ) = y n 1.) If the Wronskian is nonzero at some point t 0, then the n functions are linearly independent. (And in fact, our linearly independent solutions will have non-zero Wronskian at all points on the interval on which they exist, just as in the order two case.) Finding Roots Dealing with an order n differential equation will require solving an n-degree characteristic equation, which in general may not be easy. However, there are a few techniques that may help. Theorem: If we have a rational root p/q in lowest form to the polynomial equation a n r n + a n 1 + r n a 1 r + a 0 = 0 where the a i are integers, then p is a divisor of a 0 and q is a divisor of a n. Factor r 3 + r 4r 4 = 0. We note that any rational roots must be whole numbers which divide 4, so we have only ±1, ±, and ±4 as possibilities. Plugging in r = 1 gives ( 1) 3 + ( 1) 4( 1) 4 = 0 so r = 1 is a root. We then divide out (r + 1), to get r 3 + r 4r 4 = (r + 1)(r 4). Since r 4 = (r + )(r ), we finally have r 3 + r 4r 4 = (r + 1)(r + )(r )

3 and roots r = 1, r =, and r =. Find the roots of r 4 + 7r 3 3r 18r = 0. (I recommend working on scratch paper; you may need a few attempts.) In general however, the roots of an arbitrary degree n polynomial may be irrational, even with integer coefficients. Factoring an arbitrary polynomial can be a difficult problem. It is not uncommon to have to solve for n th roots of a number. There will be n of these in the complex plane, and can be found by writing the equation in polar form r n e nθi and using the fact that angles are only determined up to a multiple of π. Find the roots of x = 0. We have x 3 = 7, or writing x and 7 in polar form, r 3 e 3it = 7e πi. Since r is the magnitude, we can set it to 7 1/3. We then need to solve e 3ti = e πi. We have 3t = π, or t = π/3. But we might also have 3t = π + π = 3π, so t = π or even 3t = π + 4π = 5π, so t = 5π 3. After that, we reach only 3t = 7π, or t = 7π/3 which is the same as π/3. So the three roots are r 1 = (7) 1/3 e π/3i = (7) 1/3 (cos(π/3) + i sin(π/3)) = (7)1/3 r = (7) 1/3 e πi = (7) 1/3 r 3 = (7) 1/3 e 5π/3i = (7) 1/3 (cos(5π/3) + i sin(5π/3)) = (7)1/3 + i (7)1/3 3 i (7)1/3 3 3

4 See also Solving the Differential Equation y (n) ay = 0 and example 4 in section 3.1 of the text (pp ). This technique is also covered in the Stewart calculus book (5 th edition) on p. A54 of Appendix G. 3 Characteristic Equations and Fundamental Solutions To solve an order n linear homogeneous equation with constant coefficients, we can again turn to the characteristic equation. For the differential equation we have characteristic equation a n y (n) + a n 1 y (n 1) + a 1 y + a 0 y = 0 a n r n + a n 1 r n 1 + a 1 r + a 0 = 0 In general, a degree n polynomial has n roots, which may be real or complex, and which may be repeated. Everything we have done generalizes in a straightforward way, based on the roots of the characteristic equation: 1. Distinct Real Roots: Each simple real root r generates a solution of the form e rt. (Simple roots are roots which are not repeated.). Complex Roots: Each simple complex root conjugate pair a ± bi corresponds to the two solutions e at cos(bt) and e at sin(bt). 3. Repeated Roots: A real root r of multiplicity n generates the original solution e rt, as well as the n 1 additional solutions te rt, t e rt,..., t n 1 e rt. In the case of a repeated complex root pair a ± bi, we get the original solutions e at cos(bt) and e at sin(bt), plus the n 1 additional solutions te at cos(bt), te at sin(bt), t e at cos(bt), t e at sin(bt),..., t n 1 e at cos(bt), t n 1 e at sin(bt). The complete set of solutions mentioned above are linearly independent and form a fundamental set of solutions for the given differential equation. Find the general solution to y (4) 5y + 6y = 0. The characteristic equation is r 4 5r 3 + 6r = 0, which factors to r (r 5r + 6) = r (r 6)(r + 1) = 0 4

5 So we have the roots r = 0 (of multiplicity two), r = 6 and r = 1. Our fundamental set of solutions is Thus, the general solution is y 1 (t) = e 0t = 1 y (t) = te 0t = t y 3 (t) = e 6t y 4 (t) = e t y(t) = c 1 + c t + c 3 e 6t + c 4 e t Find the general solution to y (5) 3y y = 0. The characteristic equation is r 5 3r 3 r = r (r 3 3r ) = 0, so we see we have a root r = 0 of multiplicity two. To find roots of r 3 3r, we will look for rational roots, which we know must be factors of. We try ±1 and ±. We find that 1 works, and dividing out r + 1 from r 3 3r leaves us with r r = (r + 1)(r ). So finally we have factorization r 5 3r 3 r = r (r + 1) (r ) which means we have roots 0 (multiplicity two), 1 (multiplicity two), and. Thus, the general solution is y(t) = c 1 + c t + c 3 e t + c 4 te t + c 5 e t. Solve y (6) y = 0. We have characteristic equation r 6 r = 0. We can factor out an r and be left with r (r 4 1) = 0. So we now need the fourth roots of 1. We can proceed either by noting r 4 1 = (r 1)(r + 1)(r + 1) = (r 1)(r + 1)(r i)(r + i) 5

6 or by our general procedure of setting e 4ti = 1 = e 0 so that we get roots 4t = 0 (so t = 0, and e 0i = 1) 4t = π (so t = π/, and e π/ i = i) 4t = 4π (so t = π, and e πi = 1) 4t = 6π (so t = 3π/, and e 3π/ i = i) In either case, we have the following roots to our characteristic equation: r = 0 (multiplicity ), 1, 1, i, i and so we have general solution y(t) = c 1 + c t + c 3 e t + c 4 e t + c 5 cos(t) + c 6 sin(t) Find the general solution of y (4) + y + y = 0. We have characteristic equation r 4 + r + 1 = (r + 1) = 0, so we have ±i as roots, each with multiplicity two. Thus we have fundamental solutions: y 1 (t) = cos(t) y (t) = t cos(t) y 3 (t) = sin(t) y 4 (t) = t sin(t) (from repeated roots) (from repeated roots) So the general solution is y(t) = c 1 cos(t) + c t cos(t) + c 3 sin(t) + c 4 t sin(t). Find the general solution to y 8y + y 0y = 0. Characteristic equation: Roots: 6

7 Solution: 7