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

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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).

We now return to the more general equation (8),

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

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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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