Fig. 30. Solution in Example 2
2.2 Homogeneous Linear ODEs
with Constant Coefficients
EXAMPLE 2 (continued)
Initial Value Problem in the Case of Distinct Real Roots
Section 2.2 p
If the discriminant a2 − 4b is zero, we see directly from (4) that we get only one root, λ = λ1 = λ2 = −a/2, hence only one solution,
y1 = e−(a/2)x
To obtain a second independent solution y2 (needed for a basis), we use the method of reduction of order discussed in the last section, setting y2 = uy1.
Substituting this and its derivatives y’2 = u’y1 + uy'1 and y”2 into (1), we first have
(u”y1 + 2u’y’1 + uy”1) + a(u’y1 + uy’1) + buy1 = 0.
2.2 Homogeneous Linear ODEs
with Constant Coefficients
Case II. Real Double Root λ = −a/2
Section 2.2 p
Collecting terms in u”, u’, and u, as in the last section, we obtain
u”y1 + u’(2y’1 + ay1) + u(y”1 + ay’1 + by1) = 0.
The expression in the last parentheses is zero, since y1 is a solution of (1). The expression in the first parentheses is zero, too, since
2y’1 = −ae−ax/2 = −ay1.
We are thus left with u”y1 = 0. Hence u” = 0. By two integrations, u = c1x + c2.
To get a second independent solution y2 = uy1, we can simply choose c1 = 1, c2 = 0 and take u = x. Then y2 = xy1.
Since these solutions are not proportional, they form a basis.
2.2 Homogeneous Linear ODEs
with Constant Coefficients
Case II. Real Double Root λ = −a/2 (continued 1)
Section 2.2 p
Hence in the case of a double root of (3) a basis of solutions of (1) on any interval is
e−ax/2, xe−ax/2.
The corresponding general solution is
Share with your friends: |
