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Elementary Differential Equations and Boundary Value Problems - Boyce W.E.

Boyce W.E. Elementary Differential Equations and Boundary Value Problems - John Wiley & Sons, 2001. - 1310 p.
Download (direct link): elementarydifferentialequations2001.pdf
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Suppose that y1 (1) = erit and y2(1) = er2f are two solutions of an equation of the form (2). Show that they form a fundamental set of solutions if r1 = r2.
We calculate the Wronskian of y1 and y2:
W
erit
r1erit
er2t
r2er2t
= (r2 - r1) exp[(r1 + r2)1 ].
Since the exponential function is never zero, and since r2 - r1 = 0 by the statement of the problem, it follows that W is nonzero for every value of 1. Consequently, y1 and y2 form a fundamental set of solutions.
Show that y1 (1) = 11/2 and y2(t) = 1 1 form a fundamental set of solutions of
212y" + 31y' - y = 0, 1 > 0. (14)
We will show in Section 5.5 howto solve Eq. (14); see also Problem 38 in Section 3.4. However, at this stage we can verify by direct substitution that y1 and y2 are solutions of the differential equation. Since y1 (t) = 11-1/2 and y'{(1) = - 4 t-3/2,wehave
212(-11-3/2) + 31 (2 -1/2) -11/2 = (-2 +3 -1)11/2 = 0.
Similarly,y2(1) = 1-2 andy2(1) = 21-3, so
212(21-3) + 31 (-1-2) - 1-1 = (4 - 3 - 1)1-1 = 0.
Next we calculate the Wronskian W of y1 and y2:
11/2 1-1
W
21-1/2 -t-2
= - 1-3/2. (15)
Since W = 0 for 1 > 0, we conclude that y2 and y2 form a fundamental set of solutions there.
In several cases, we have been able to find a fundamental set of solutions, and therefore the general solution, of a given differential equation. However, this is often a difficult task, and the question may arise as to whether or not a differential equation of the form (2) always has a fundamental set of solutions. The following theorem provides an affirmative answer to this question.
144
Chapter 3. Second Order Linear Equations
Theorem 3.2.5
EXAMPLE
6
Consider the differential equation (2),
L [y] = y + p(1) + q (1) = 0,
whose coefficients p and q are continuous on some open interval I. Choose some point t0 in I. Let y2 be the solution ofEq. (2) that also satisfies the initial conditions
y%) = 1, (10) = 0,
and let y2 be the solution of Eq. (2) that satisfies the initial conditions
y(10) = 0, %) = 1.
Then y2 and y2 form a fundamental set of solutions of Eq. (2).
First observe that the existence of the functions y2 and y2 is assured by the existence part of Theorem 3.2.1. To show that they form a fundamental set of solutions we need only calculate their Wronskian at t0:
W (ӳ, 2)(0 ) =
1 0
0 1
1.
yi(10) y2(10)
Ӳ (10) 2 (10)
Since their Wronskian is not zero at the point t0, the functions y2 and y2 do form a fundamental set of solutions, thus completing the proof of Theorem 3.2.5.
Note that the difficult part of this proof, demonstrating the existence of a pair of solutions, is taken care of by reference to Theorem 3.2.1. Note also that Theorem 3.2.5 does not address the question of how to solve the specified initial value problems so as to find the functions y1 and y2 indicated in the theorem. Nevertheless, it may be reassuring to know that a fundamental set of solutions always exists.
Find the fundamental set of solutions specified by Theorem 3.2.5 for the differential equation
' - = 0, (16)
using the initial point t0 = 0.
In Section 3.1 we noted that two solutions ofEq. (16) are y2(1) = ef and y2(1) =
e-1. The Wronskian of these solutions is W(y2,y2)(1) = -2 = 0, so they form a
fundamental set of solutions. However, they are not the fundamental solutions indicated by Theorem 3.2.5 because they do not satisfy the initial conditions mentioned in that theorem at the point 1 = 0.
To find the fundamental solutions specified by the theorem we need to find the solutions satisfying the proper initial conditions. Let us denote by y3(1) the solution of Eq. (16) that satisfies the initial conditions
y(0) = 1, y'(0) = 0. (17)
The general solution ofEq. (16) is
y = c2e + c2e-1, (18)
and the initial conditions (17) are satisfied if c2 = 1/2 and c2 = 1/2. Thus
y3(t) = 2 e1 + 2 e-1 = cosh 1.
3.2 Fundamental Solutions of Linear Homogeneous Equations
145
Similarly, if y4(t) satisfies the initial conditions
y(0) = 0, y'(0) = 1, (19)
then
y4(t) = 2e1 - 2e-1 = sinh 1.
Since the Wronskian of y3 and y4 is
W(y3,yA)(1) = cosh21 - sinh21 = 1,
these functions also form a fundamental set of solutions, as stated by Theorem 3.2.5. Therefore, the general solution ofEq. (16) can be written as
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