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Elementary differential equations 7th edition - Boyce W.E

Boyce W.E Elementary differential equations 7th edition - Wiley publishing , 2001. - 1310 p.
ISBN 0-471-31999-6
Download (direct link): elementarydifferentialequat2001.pdf
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The next theorem justifies the term “general solution” that we introduced in Section 3.1 for the linear combination c1 y1 + c2y2.
Theorem 3.2.4 If y1 and y2 are two solutions of the differential equation (2),
L [y] = y” + p(t ) y + q (t ) y = 0,
and if there is a point t0 where the Wronskian of y1 and y2 is nonzero, then the family of solutions
y = c1 y1 (t ) + c2 y2(t )
with arbitrary coefficients c1 and c2 includes every solution of Eq. (2).
Let 0 be any solution of Eq. (2). To prove the theorem we must show that 0 is included in the linear combination c1 y1 + c2y2; that is, for some choice of the constants c1 and c2, the linear combination is equal to 0. Let t0 be a point where the Wronskian of y1 and y2 is nonzero. Then evaluate 0 and 0' at this point and call these values y0 and y0, respectively; thus
y0 = 0(t0), y0 = 0'(t0)-
Next, consider the initial value problem
y" + p(t)y + q (t)y = o, y(t0) = y0, y'(t0) = y0 • (13)
The function 0 is certainly a solution of this initial value problem. On the other hand, since W(y1,y2)(t0) is nonzero, it is possible (by Theorem 3.2.3) to choose c1 and c2 so that y = c1 y1 (t) + c2y2(t) is also a solution of the initial value problem (13). In fact, the proper values of c1 and c2 are given by Eqs. (10) or (11). The uniqueness part of Theorem 3.2.1 guarantees that these two solutions of the same initial value problem are actually the same function; thus, for the proper choice of c1 and c2,
0(t ) = c1 y1(t ) + c2y2(t ),
and therefore 0 is included in the family of functions of c1 y1 + c2y2. Finally, since 0 is an arbitrary solution of Eq. (2), it follows that every solution of this equation is included in this family. This completes the proof of Theorem 3.2.4.
Theorem 3.2.4 states that, as long as the Wronskian of y1 and y2 is not everywhere zero, the linear combination c1 y1 + c2y2 contains all solutions of Eq. (2). It is therefore natural (and we have already done this in the preceding section) to call the expression
y = c1 y1(t ) + c2y2(t )
with arbitrary constant coefficients the general solution of Eq. (2). The solutions y1 and y2, with a nonzero Wronskian, are said to form a fundamental set of solutions of Eq. (22).
3.2 Fundamental Solutions of Linear Homogeneous Equations
143
EXAMPLE
4
EXAMPLE
5
To restate the result of Theorem 3.2.4 in slightly different language: To find the general solution, and therefore all solutions, of an equation of the form (2), we need only find two solutions of the given equation whose Wronskian is nonzero. We did precisely this in several examples in Section 3.1, although there we did not calculate the Wronskians. You should now go back and do that, thereby verifying that all the solutions we called “general solutions” in Section 3.1 do satisfy the necessary Wronskian condition. Alternatively, the following example includes all those mentioned in Section 3.1, as well as many other problems of a similar type.
Suppose that y1(t ) = eW and y2(t ) = eW 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)t].
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 t. Consequently, y1 and y2 form a fundamental set of solutions.
Show that y1 (t) = t1/2 and y2(t) = t 1 form a fundamental set of solutions of
2t V + 3ty' — y = 0, t > 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'{(t) = — 4 t-3/2,wehave
2t 2(—41—3/2) + 3t ( 21—1/2) — t1/2 = (—2 +3 — 1)t1/2 = o.
Similarly,y2(t) = —t—2 andy2(t) = 2t—3, so
2t 2(2t—3) + 3t (—t—2) — t—1 = (4 — 3 — 1)t—1 = 0.
Next we calculate the Wronskian W of y1 and y2:
t1/2 t—1
W
21—1/2 —1—2
= -§ t—3/2. (15)
Since W = o for t > o, we conclude that y1 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(t) / + q (t) y = 0,
whose coefficients p and q are continuous on some open interval I. Choose some point t0 in I. Let y1 be the solution of Eq. (2) that also satisfies the initial conditions
y(to) = 1, / (t0) = 0,
and let y2 be the solution of Eq. (2) that satisfies the initial conditions
y(t0) = 0, y'(t0) = 1.
Then y1 and y2 form a fundamental set of solutions of Eq. (2).
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