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
2. 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 p0(t), p1(t), . . . , pn−1(t) and g(t) be
continuous functions defined on a < t < b, and let t0 be in (a, b). Then
the initial value problem
y(n)+ pn−1(t)y(n−1)+ · · · + p2(t)y′′+ p1(t)y′+ p0(t)y = g(t)
y(t0) = y0, y′(t0) = y′0, y′′(t0) = y′′0, . . . , y(n−1)(t0) = y0(n−1)
has a unique solution defined on the entire interval (a, b).
So suppose we have an nth order linear homogeneous differential equation
y(n)(t) + p1(t)y(n−1)(t) + p2(t)y(n−2)(t) + · · · + pn(t)y(t) = 0
defined on some interval I where the piare continuous and we are therefore guaranteed
a solution. Then we will need n fundamental solutions y1, . . . , yn, which are to be
linearly independent. In that case, we have a general solution y(t) = c1y1(t) + c2y2(t) + · · · cnyn(t).
To check that our proposed fundamental solutions yi are in fact linearly independent,
we use the Wronskian, which now takes the form
W (y1, y2, . . . yn)(t) = y1(t) y2(t) · · · yn(t) y′ 1(t) y2′(t) · · · yn′(t) ... ... . .. ... y1(n−1)(t) y2(n−1)(t) · · · y(n−1) n (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(t0) = y0through y(n−1)(t0) = yn−1.) If the
Wronskian is nonzero at some point t0, 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.)
2
Finding Roots
Dealing with an order n differential equation will require solving an n-degree char-acteristic 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 poly-nomial equation
anrn+ an−1+ rn−1+ · · · + a1r + a0 = 0
where the ai are integers, then p is a divisor of a0 and q is a divisor of an.
Example: Factor r3+ r2
− 4r − 4 = 0.
We note that any rational roots must be whole numbers which divide 4, so we have only ±1, ±2, and ±4 as possibilities. Plugging in r = −1 gives
(−1)3+ (−1)2− 4(−1) − 4 = 0 so r = −1 is a root. We then divide out (r + 1), to get
r3+ r2− 4r − 4 = (r + 1)(r2− 4). Since r2
− 4 = (r + 2)(r − 2), we finally have
and roots r = −1, r = −2, and r = 2. Example:
Find the roots of 2r4+ 7r3
− 3r2− 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 nth 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 rnenθi and using the fact that angles are only determined up to a multiple of 2π.
Example:
Find the roots of x3+ 7 = 0.
We have x3 = −7, or writing x and −7 in polar form, r3e3it = 7eπi. Since r is the
magnitude, we can set it to 71/3. We then need to solve
e3ti = eπi. We have 3t = π, or t = π/3. But we might also have
3t = π + 2π = 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
r1 = (7)1/3eπ/3i = (7)1/3(cos(π/3) + i sin(π/3)) =
(7)1/3
2 + i
(7)1/3√3
2 r2 = (7)1/3eπi = −(7)1/3
r3 = (7)1/3e5π/3i = (7)1/3(cos(5π/3) + i sin(5π/3)) =
(7)1/3
2 − i
(7)1/3√3
See also “Solving the Differential Equation y(n)
− ay = 0 and example 4 in section 3.12 of the text (pp. 198-9). This technique is also covered in the Stewart calculus book (5th edition) on p. A54 of Appendix G.
3
Characteristic Equations and Fundamental
So-lutions
To solve an order n linear homogeneous equation with constant coefficients, we can again turn to the characteristic equation. For the differential equation
any(n)+ an−1y(n−1)+ · · · a1y′+ a0y = 0
we have characteristic equation
anrn+ an−1rn−1+ · · · a1r + a0 = 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 ert. (Simple roots are roots which are not repeated.)
2. Complex Roots: Each simple complex root conjugate pair a ± bi corresponds to the two solutions eatcos(bt) and eatsin(bt).
3. Repeated Roots: A real root r of multiplicity n generates the original solu-tion ert, as well as the n − 1 additional solutions tert, t2ert, . . ., tn−1ert. In
the case of a repeated complex root pair a ± bi, we get the original solu-tions eatcos(bt) and eatsin(bt), plus the n − 1 additional solutions teatcos(bt),
teatsin(bt), t2eatcos(bt), t2eatsin(bt), . . ., tn−1eatcos(bt), tn−1eatsin(bt).
The complete set of solutions mentioned above are linearly independent and form a fundamental set of solutions for the given differential equation.
Example:
Find the general solution to y(4)
− 5y′′′+ 6y′′ = 0.
The characteristic equation is r4− 5r3+ 6r2 = 0, which factors to
So we have the roots r = 0 (of multiplicity two), r = 6 and r = −1. Our fundamental set of solutions is y1(t) = e0t= 1 y2(t) = te0t = t y3(t) = e6t y4(t) = e−t
Thus, the general solution is
y(t) = c1 + c2t + c3e6t+ c4e−t
Example:
Find the general solution to y(5)
− 3y′′′− 2y′′= 0.
The characteristic equation is r5− 3r3− 2r2 = r2(r3− 3r − 2) = 0, so we see we have
a root r = 0 of multiplicity two. To find roots of r3
− 3r − 2, we will look for rational roots, which we know must be factors of 2. We try ±1 and ±2. We find that −1 works, and dividing out r + 1 from r3− 3r − 2 leaves us with r2− r − 2 = (r + 1)(r − 2). So finally we have factorization
r5− 3r3− 2r2 = r2(r + 1)2(r − 2)
which means we have roots 0 (multiplicity two), −1 (multiplicity two), and 2. Thus, the general solution is
y(t) = c1+ c2t + c3e−t+ c4te−t+ c5e2t.
Example: Solve y(6)
− y′′ = 0.
We have characteristic equation r6
− r2 = 0. We can factor out an r2 and be left with
r2(r4− 1) = 0.
So we now need the fourth roots of 1. We can proceed either by noting r4− 1 = (r − 1)(r + 1)(r2+ 1) = (r − 1)(r + 1)(r − i)(r + i)
or by our general procedure of setting e4ti = 1 = e0 so that we get roots
4t = 0 (so t = 0, and e0i = 1) 4t = 2π (so t = π/2, and eπ/2 i= i) 4t = 4π (so t = π, and eπi = −1) 4t = 6π (so t = 3π/2, and e3π/2 i = −i)
In either case, we have the following roots to our characteristic equation: r = 0 (multiplicity 2), 1, −1, i, −i
and so we have general solution
y(t) = c1+ c2t + c3et+ c4e−t+ c5cos(t) + c6sin(t)
Example:
Find the general solution of y(4)+ 2y′′+ y = 0.
We have characteristic equation r4+ 2r2+ 1 = (r2+ 1)2 = 0, so we have ±i as roots,
each with multiplicity two.
Thus we have fundamental solutions: y1(t) = cos(t)
y2(t) = t cos(t) (from repeated roots)
y3(t) = sin(t)
y4(t) = t sin(t) (from repeated roots)
So the general solution is
y(t) = c1cos(t) + c2t cos(t) + c3sin(t) + c4t sin(t).
Example:
Find the general solution to y′′′− 8y′′+ 22y′− 20y = 0.
Characteristic equation: Roots:
