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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
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Equation (16) is called the characteristic equation for the differential equation (8). Its significance lies in the fact that if r is a root of the polynomial equation (16), then y = ert is a solution of the differential equation (8). Since Eq. (16) is a quadratic equation with real coefficients, it has two roots, which may be real and different, real but repeated, or complex conjugates. We consider the first case here, and the latter two cases in Sections 3.4 and 3.5.
Assuming that the roots of the characteristic equation (16) are real and different, let them be denoted by r1 and r2, where r1 = r2. Then y1(t) = er1t and y2(t) = er2( are two solutions of Eq. (8). Just as in the preceding example, it now follows that
is also a solution of Eq. (8). To verify that this is so, we can differentiate the expression in Eq. (17); hence
Substituting these expressions for y, y;, and y" in Eq. (8) and rearranging terms, we obtain
ay" + by' + cy = c1(ar1 + br1 + c)er1t + c2(ar| + br2 + c)er2t. (20)
ay" + by' + cy = 0,
r is a parameter to be determined. Then it follows that y' = rert and y" = r2ert. By substituting these expressions for y, y;, and y" in Eq. (8), we obtain
(ar2 + br + c)ert = 0,
or, since ert = 0,
ar2 + br + c = 0.
(16)
y = c1 y1(t ) + c2y2(t ) = c1er1t + c2er2t
(17)
y' = c1r1er1t + c2r2er2t
(18)
and
(19)
3.1 Homogeneous Equations with Constant Coefficients
133
The quantity in each of the parentheses on the right side of Eq. (20) is zero because r1
and r2 are roots of Eq. (16); therefore, y as given by Eq. (17) is indeed a solution of
Eq. (8), as we wished to verify.
Now suppose that we want to find the particular member of the family of solutions (17) that satisfies the initial conditions (6),
y (t0) = y0’ y'(t0) = y0 •
By substituting t = t0 and y = y0 in Eq. (17), we obtain
c1er1t0 + c2er2t0 = y0. (21)
Similarly, setting t = t0 and y' = y0 in Eq. (18) gives
c1r1er1t0 + c2r2er2t0 = y0 • (22)
On solving Eqs. (21) and (22) simultaneously for c1 and c2, we find that
e-Vo, , = yori - y e-^o. (23)
Thus, no matter what initial conditions are assigned, that is, regardless of the values of t0, y0, and y0 in Eqs. (6), it is always possible to determine c1 and c2 so that the initial conditions are satisfied; moreover, there is only one possible choice of c1 and c2 for each set of initial conditions. With the values of c1 and c2 given by Eq. (23), the expression (17) is the solution of the initial value problem
ay" + by'+ cy = 0’ y(t0) = y0’ y'(t0) = y0 • (24)
It is possible to show, on the basis of the fundamental theorem cited in the next section, that all solutions of Eq. (8) are included in the expression (17), at least for the case in which the roots of Eq. (16) are real and different. Therefore, we call Eq. (17) the general solution of Eq. (8). The fact that any possible initial conditions can be satisfied by the proper choice of the constants in Eq. (17) makes more plausible the idea that this expression does include all solutions of Eq. (8).
EXAMPLE
1
Find the general solution of
y" + 5 y' + 6 y = 0. (25)
We assume that y = ert, and it then follows that r must be a root of the characteristic
equation
r2 + 5r + 6 = (r + 2)(r + 3) = 0.
Thus the possible values of r are r1 = —2 and r2 = — 3; the general solution of
Eq. (25) is 1 2
y = c1e—2t + c2e~3t • (26)
Find the solution of the initial value problem
IEXAMPT E
2 y" + 5 y' + 6 y = 0, y(0) = 2, y'(0) = 3. (27)
134
Chapter 3. Second Order Linear Equations
EXAMPLE
3
The general solution of the differential equation was found in Example 1 and is given by Eq. (26). To satisfy the first initial condition we set t = 0 and y = 2 in Eq. (26); thus c1 and c2 must satisfy
c1 + c2 = 2. (28)
To use the second initial condition we must first differentiate Eq. (26). This gives y' = —2c1e~2t - 3c2e-3t. Then, setting t = 0 and y' = 3, we obtain
-2c1 - 3c2 = 3. (29)
By solving Eqs. (28) and (29) we find that c1 = 9 and c2 = —7. Using these values in the expression (26), we obtain the solution
y = 9e-2t - 7e-3t (30)
of the initial value problem (27). The graph of the solution is shown in Figure 3.1.1.
FIGURE 3.1.1 Solution of y" + 5y' + 6y = 0, y(0) = 2, y'(0) = 3.
Find the solution of the initial value problem
4y"- 8y' + 3y = 0, y(0) = 2, /(0) = 1. (31)
If y = ert, then the characteristic equation is
4r2 - 8r + 3 = 0
and its roots are r = 3/2andr = 1/2. Therefore the general solution of the differential equation is
y = cxe2t/2 + c2et/2. (32)
Applying the initial conditions, we obtain the following two equations for cx and c2:
C1 + c2 = 2> 2c1 + 2c2 = 2 '
The solution of these equations is cx = — 1, c2 = 5, and the solution of the initial value problem (31) is
y = — e3t/2 + 2 et/2. (33)
Figure 3.1.2 shows the graph of the solution.
3.1 Homogeneous Equations with Constant Coefficients
135
EXAMPLE
4
FIGURE 3.1.2 Solution of 4y" — 8y' + 3y = 0, y(0) = 2, y (0) = 0.5.
The solution (30) of the initial value problem (27) initially increases (because its initial slope is positive) but eventually approaches zero (because both terms involve negative exponential functions). Therefore the solution must have a maximum point and the graph in Figure 3.1.1 confirms this. Determine the location of this maximum point.
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