Download Homogeneous Equations with Constant Coefficients

Jim Lambers
MAT 285
Spring Semester 2012-13
Lecture 15 Notes
These notes correspond to Section 3.1 in the text.
Homogeneous Equations with Constant Coefficients
We have learned various techniques for solving first-order ODE, that only involve the first derivative
of the solution. We now consider second-order ODE. The most general second-order ODE is an
equation of the form
F (t, y, y 0 , y 00 ) = 0
for some function F of 4 variables. Such an equation can be either linear or nonlinear. We will
focus on linear second-order equations of the form
y 00 + p(t)y 0 + q(t)y = g(t),
with coefficients p(t) and q(t) and a right-hand side g(t). If g(t) = 0, we say that the equation is
homogeneous (not to be confused with homogeneous first-order equations considered previously);
otherwise, it is said to be inhomogeneous. If p(t) and q(t) are constants, we say that the equation
is constant-coefficient; otherwise, it is variable-coefficient.
We now consider how to solve a linear homogeneous second-order ODE with constant coefficients,
y 00 + py 0 + qy = 0.
Such an equation is normally part of an initial value problem that has two initial conditions,
y 0 (0) = z0 ,
y(0) = y0 ,
due to the second derivative in the ODE. In order to solve this problem, we can leverage our
knowledge of first-order linear equations.
First, we define the variables
u1 = y, u2 = y 0 .
Then our second-order ODE can be rewritten as a system of first-order ODE,
u01 = u2 ,
u02 + pu2 + qu1 = 0,
with accompanying initial conditions
u1 (0) = y0 ,
u2 (0) = z0 .
This initial value problem can be expressed in matrix-vector form as follows:
u0 = Au,
where
u=
u1
u2
,
A=
u0 (t0 ) = u0 ,
0
1
−q −p
1
,
u(t0 ) =
y0
z0
.
Recall that the solution of the first-order initial value problem
y 0 = ay,
y(0) = y0
is y(t) = eat y0 . Similarly, the solution of the preceding first-order system is
u(t) = eAt u0
where the matrix exponential eAt is defined by
e
At
=
∞
X
(At)j
j!
j=0
,
based on the Maclaurin series for the exponential function
ex =
∞
X
xj
j=0
j!
.
However, this form of the solution is not convenient for describing the solution y(t) of our original
second-order problem, so we must use an alternative approach.
To that end, we consider the question of when the system u0 = Au, for a general 2 × 2 matrix
A, is easy to solve. This is certainly the case when A is a diagonal matrix of the form
λ1 0
.
A=
0 λ2
In this case, the equations in the system,
u01 = λ1 u1 ,
u02 = λ2 u2 ,
are decoupled, and therefore can be solved independently. This yields the solutions
u1 (t) = eλ1 t y0 ,
u2 (t) = eλ2 t z0 .
Therefore, we consider whether our system that is obtained from a second-order ODE can somehow
be transformed into a system in which the equations are decoupled.
We define new variables w1 and w2 through a linear transformation of u1 and u2 :
u1 = p11 w1 + p12 w2 ,
u2 = p21 w1 + p22 w2 ,
where pij is a constant for i, j = 1, 2. This transformation can be expressed in matrix-vector form
as
p11 p12
w1
u = P w, P =
, w=
.
p21 p22
w2
Substituting u = P w into the system u0 = Au yields
(P w)0 = P w0 + P 0 w = AP w.
Since the entries of P are constants, P 0 = 0. Multiplying both sides on the left by P −1 yields
w0 = Dw,
D = P −1 AP.
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Our goal is to choose P so that D is a diagonal matrix with diagonal entries λ1 and λ2 , as before.
Multiplying both sides of D = P −1 AP on the left by P yields AP = P D. If we examine each
column of this equation, where D is a diagonal matrix, we obtain the equations
Api = λi pi ,
i = 1, 2,
where pi is the ith column of P . The vector pi is an eigenvector of A, and λi is an eigenvalue. For
P to be invertible, pi must be a nonzero vector, and therefore the system of equations
(A − λi I)pi = 0,
where I is the identity matrix, must have a nonzero solution. It follows that we must have
det(A − λi I) = 0.
We therefore examine det(A − λI), where λ is a variable. We have
−λ
1
det(A − λI) = det
= −λ(−p − λ) − q = λ2 + pλ + q.
−q −p − λ
It follows that λ1 and λ2 are the two solutions of the quadratic equation
λ2 + pλ + q = 0.
This equation is called the characteristic equation of the ODE y 00 + py 0 + q = 0. Note that the
characteristic equation is obtained by replacing differentiation with powers of λ.
Let λ1 , λ2 be the roots of the characteristic equation, which can be obtained using the quadratic
formula. For now, we assume that these roots are real and distinct. Then, we need to obtain the
vectors p1 and p2 , by solving the equations
−λi
1
p1i
0
, i = 1, 2.
=
−q −p − λi
p2i
0
Using the first equation in this system, we obtain the solutions
1
, i = 1, 2.
pi =
λi
That is,
P =
1 1
λ1 λ2
,
which yields
P
−1
1
=
λ2 − λ1
It follows that
−1
eAt = eP DP t
∞
X
1
=
(P DP −1 t)j
j!
j=0
∞
X
= P
j=0
1
(Dt)j P −1
j!
3
λ2 −1
−λ1 1
.
= P eDt P −1
1
=
λ2 − λ1
1
=
λ2 − λ1
1
=
λ2 − λ1
e λ1 t
0
λ2 −1
−λ1 1
0 eλ2 t
λ
t
λ
t
e 1
e 2
λ2 −1
−λ1 1
λ1 eλ1 t λ2 eλ2 t
λ
t
λ
t
1
2
λ2 e − λ1 e
eλ2 t − eλ1 t
λ1 λ2 (eλ1 t − eλ2 t ) λ2 eλ2 t − λ1 eλ1 t
1 1
λ1 λ2
Therefore the solution of our original second-order initial value problem is
y(t) = u1 (t)
1 0 u(t)
=
1 0 eAt u0
=
y0
1
λ
t
λ
t
λ
t
λ
t
1
2
2
1
λ2 e − λ1 e
e −e
=
z0
λ2 − λ1
=
=
eλ2 t − eλ1 t
λ 2 e λ1 t − λ 1 e λ2 t
y0 +
z0
λ2 − λ1
λ2 − λ1
λ2 y0 − z0 λ1 t z0 − λ1 y0 λ2 t
e +
e .
λ2 − λ1
λ2 − λ1
It can be seen that the solution y(t) is a linear combination of the functions eλ1 t and eλ2 t , where
λ1 and λ2 are the roots of the characteristic equation.
This observation leads to the following practical method for solving a second-order linear homogeneous ODE with constant coefficients. First, the characteristic equation is solved for λ1 and
λ2 . If these roots are real and distinct, then the function
y(t) = c1 eλ1 t + c2 eλ2 t
is the general solution of the ODE. The constants c1 and c2 can then be obtained by substituting
the general solution into the initial conditions. Later, we will consider the case where the roots are
either repeated or complex.
Example We will solve the initial value problem
y 00 + 5y 0 + 6y = 0,
y(0) = 2,
y 0 (0) = −1.
The characteristic equation is
λ2 + 5λ + 6 = 0.
Factoring yields (λ + 2)(λ + 3) = 0, and the roots λ1 = −2, λ2 = −3. Thus the general solution is
y(t) = c1 e−2t + c2 e−3t .
Substituting the general solution into the initial conditions yields the equations
y(0) = 2 = c1 + c2 ,
y 0 (0) = −1 = −2c1 − 3c2 .
Solving these equations simultaneously, we obtain c1 = 5 and c2 = −3. We conclude that the
solution of the initial value problem is
y(t) = 5e−2t − 3e−3t .
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