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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.
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Let us summarize what we have done so far in this example. Once we notice that the functions y1 (t) = º* and y2(t) = e~( are solutions ofEq. (9), it follows that the general linear combination (10) of these functions is also a solution. Since the coefficients c1 and c2 in Eq. (10) are arbitrary, this expression represents a doubly infinite family of solutions of the differential equation (9).
It is now possible to consider how to pick out a particular member of this infinite family of solutions that also satisfies a given set of initial conditions. For example, suppose that we want the solution of Eq. (9) that also satisfies the initial conditions
y(0) = 2, y'(0) = -1. (11)
In other words, we seek the solution that passes through the point (0, 2) and at that point has the slope -1. First, we set t = 0 and y = 2 in Eq. (10); this gives the equation
c1 + c2 = 2.
(12)
132
Chapter 3. Second Order Linear Equations
Next, we differentiate Eq. (10) with the result that
tt y = c1e - c2e .
Then, setting t = 0 and y' = -1, we obtain
(13)
By solving Eqs. (12) and (13) simultaneously for c1 and c2 we find that
c - 1 c - 3
C1 = 2 ’ 2 = 2.
Finally, inserting these values in Eq. (10), we obtain
(14)
(15)
the solution of the initial value problem consisting of the differential equation (9) and the initial conditions (11).
which has arbitrary (real) constant coefficients. Based on our experience with Eq. (9), let us also seek exponential solutions ofEq. (8). Thus we suppose that y = ert, where
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) = e^ are two solutions ofEq. (8). Just as in the preceding example, it now follows that
is also a solution ofEq. (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(ar2 + br2 + c)er2. (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 + c2er2
(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 ofEq. (20) is zero because r1
and r2 are roots ofEq. (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),
Ó (t0) = Ó0> y'(t0) = Ó0.
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-r1t0, c2 = y0r1 - y0 e-r2t0. (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 = °, y(t0) = Ó0, y'(t0) = y0. (24)
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