Jim Lambers
MAT 285
Spring Semester 2016-17
Lecture 14 Example
These notes correspond to Section 3.1 in the text.
Second-Order Linear Equations with Constant Coefficients
Consider the second-order, linear, homogeneous equation with constant coefficients
y 00 + py 0 + qy = 0,
(1)
with initial conditions
y(0) = y0 ,
y 0 (0) = z0 .
(2)
u2 = y 0
(3)
By introducing the variables
u1 = y,
equation (1) can be expressed as a system of two first-order equations
u01 = u2 ,
(4)
u02
(5)
= −qu1 − pu2 .
Unfortunately, these equations cannot be solved independently of one another, because both equations involve both unknowns u1 and u2 . We therefore seek a change of variable of the form
u1 = p11 w1 + p12 w2 ,
(6)
u2 = p21 w1 + p22 w2 ,
(7)
where pij is a constant for i, j = 1, 2, so that w10 will depend only on w1 , and w20 will depend only
on w2 .
Substituting equations (6), (7) into equations (4), (5) yields
p11 w10 + p12 w20 = p21 w1 + p22 w2 ,
(8)
p21 w10
(9)
+
p22 w20
= −q(p11 w1 + p12 w2 ) − p(p21 w1 + p22 w2 ).
We then rearrange equation (8) to isolate w1 ’, which yields
w10 =
1
[p21 w1 + p22 w2 − p12 w20 ].
p11
(10)
Substituting this expression for w10 into equation (9) yields
p21
[p21 w1 + p22 w2 − p12 w20 ] + p22 w20 = −q(p11 w1 + p12 w2 ) − p(p21 w1 + p22 w2 ).
p11
(11)
We then multiply both sides by p11 to obtain
p21 [p21 w1 + p22 w2 − p12 w20 ] + p11 p22 w20 = −q(p211 w1 + p11 p12 w2 ) −
p(p11 p21 w1 + p11 p22 w2 ).
1
(12)
Rearranging to isolate w20 yields
w20 = −
(p221 + pp11 p21 + qp211 )w1 + (p21 p22 + qp11 p12 + pp11 p22 )w2
.
p11 p22 − p21 p12
(13)
Note that we must have the denominator p11 p22 − p21 p12 nonzero. This is the determinant of the
coefficient matrix of the system (6), (7), which means that the change of variables from u1 , u2 to
w1 , w2 must be invertible.
We want this equation to not include w1 , which means that the coefficient of w1 must be zero.
That is, we must have
p221 + pp11 p21 + qp211 = 0.
(14)
If we assume p11 6= 0, we can divide by p211 to obtain the quadratic equation
λ21 + pλ1 + q = 0,
(15)
where λ1 = p21 /p11 . That is, λ1 must be a root of the characteristic equation
λ2 + pλ + q = 0.
(16)
This condition does not specify either p11 or p21 , but it does specify their ratio. Therefore, we
assume p11 = 1, which means p21 = λ1 . Equation (13) then simplifies to
w20 = −
λ1 p22 + qp12 + pp22
w2 .
p22 − λ1 p12
(17)
Let λ2 be the other root of equation (16). Then, it follows that
λ1 + λ2 = −p,
λ1 λ2 = q.
(18)
Substituting these relations into equation (17) and simplifying yields
w20 = λ2 w2 .
(19)
Going back to equations (8), (9) and proceeding with the opposite steps, we obtain the similar
equation
w10 = λ1 w1 ,
(20)
as well as p12 = 1, p22 = λ2 .
Equations (19), (20) have the solutions
w1 (t) = w1 (0)eλ1 t ,
w2 (t) = w2 (0)eλ2 t .
(21)
From equations (3) and (6), we conclude that the general solution of equation (1) has the form
y(t) = u1 (t) = p11 w1 (t) + p12 w2 (t) = w1 (0)eλ1 t + w2 (0)eλ2 t ,
(22)
p11 p22 − p12 p21 = λ2 − λ1 6= 0.
(23)
provided that
That is, the roots of the characteristic equation (16) must be distinct. We say that y(t) is a linear
combination of the functions eλ1 t and eλ2 t .
2
We conclude by considering the initial conditions (2). By evaluating equations (6), (7) at t = 0,
we obtain
y0 = w1 (0) + w2 (0),
(24)
z0 = λ1 w1 (0) + λ2 w2 (0),
(25)
From equation (24), we obtain w2 (0) = y0 − w1 (0), which can be substituted into equation (25) to
obtain
z0 = λ1 w1 (0) + λ2 (y0 − w1 (0)),
(26)
which yields
w1 (0) =
z0 − λ2 y0
,
λ1 − λ2
w2 (0) =
λ 1 y0 − z 0
.
λ1 − λ2
(27)
In conclusion, the solution of the initial value problem defined by equations (1), (2) is
y(t) =
z0 − λ2 y0 λ1 t λ1 y0 − z0 λ2 t
e +
e ,
λ1 − λ2
λ1 − λ2
(28)
where λ1 , λ2 are the roots of the characteristic equation (16), provided that these roots are distinct.
This solution can also be obtained from the general solution from equation (22), rewritten as
y(t) = c1 eλ1 t + c2 eλ2 t ,
and then using the initial conditions (2) to solve for the coefficients c1 and c2 .
3
(29)