## Linear Second Order Differential Equations

Until now we have confined our analysis of differential equations 10 those of the first order. In this chapter we will examine linear, second-order differential equations with constant coefficients. We focus our attention on the autonomous case in section 23.1 and consider a special nonautonamous case in section 23.2.

23.1 The Linear, Autonomous,

Second-Order Differential Equation

We begin by explaining how to solve a linear, autonomous, second-order differential equation.

The linear, autonomous, second-order differential equation (constant coefficients and a constant term) is expressed as y -f- a\y -+- aiy = b (23.1)

Equation (23.1 ) is linear because y, v. and v are not raised to any power other than one, It is autonomous because it has constant coefficients, u\ and and a constant term. b. If the coefficients or the term vary with t, then the equation is nonautonomous. In section 23.2 we consider the case of a variable term.

Rather than try to solve the complete equation in one step, we exploit the fact that the complete solution to a linear differential equation is equal to the sum of the solution to its homogeneous form and a particular solution to the complete equation. In symbols y = y h + >y (23.2)

where y is the complete solution, >/, is the solution to the homogeneous form, and yr is a particular solution. Many readers are probably familiar with this technique because we used it in earlier chapters to solve linear, first-order differential equations and linear difference equations.

Definition 23.2

The General Solution to the Homogeneous Equation

The first step in solving the linear. autonomous, second-order differential equation is 10 solve the homogeneous form of equation (23.1).

The homogeneous form of the linear, second-order differential equation with constant coefficients is y 4- aijH- aiy = 0 (23.3)

To solve this second-order differential equation, we shall make use of what we already know about the solution to m first-order counterpart, the linear, homogeneous, first-order differential equation with a constant coefficient. In chapter 21 we learned that solutions to equations of this kind are of the form y(t) = Aerl (23.4)

where the values for A and r are determined by initial conditions and the coefficient of the equation. A reasonable hypothesis is that solutions to second-order equations are of the same form. Specifically, let us conjecture that equation (23.4) is a solution to equation (23.3). If we are right, then equation (23.4) must satisfy equation (23.3). To see if it does, differentiate equation (23.4) to get y = rAe" (23.5)

and differentiate again to get y = r2Aer' (23.6)

Substitute the hypothesized solution and its derivatives, equations (23.4) to (23.6), into the left-hand side of equation (23.3) and check that it satisfies the equality in equation (23.3). The left-hand side becomes v 4- tt\y 4- tij.V = r2Ae" 4- u\rAert 4- a2Ae" = Aen(r1 4-a,r 4- Ui)

Ruling out the special (and trivial) case of A = 0. our conjecture is correct it the expression in brackets is equal to zero, for then our conjectured solution satisfies equation (23.3). But since r is, as yet, an unspecified parameter in the solution, we are free to choose it to make the expression in brackets identically equal to zero.

lii other words, if we choose r to satisfy

then equation (23.4) is indeed a solution to equation (23.3).

Equation (23.7). known as the characteristic equation, plays an important role in finding the solution to equation (23.3).

Definition 23.3

The characteristic equation of the linear second-order differential equation with constant coefficients is r' + 0\r + a2 = 0

The values of r that solve the characteristic equation are known as the characteristic roots (or just the roots) or eigenvalues of the characteristic equation. Since the characteristic equation is a quadratic in the case of a second-order differential equation, there are two roots: we will call these r| and r2. Their values are

If the two roots that solve the characteristic equation are different, we actually have two different solutions to equation (23.3). They are yi = Aie'1' and y2 = A2e'-' (23.9)

Let us pause to verify that each of these is a solution to equation (23.3). We do this for the first solution, V|, and leave it to the reader to verify that y2 is also a solution. Substitute yi and its first and second derivatives into equation (23.3) and check that equation (23.3) is then satisfied (i.e., check that the left-hand side does equal zero). The first derivative is y,=nAie'"

and the second derivative is

After making substitutions, the left-hand side of equation (23.3) becomes

.y-t-a,y + a2.v = r\A\er>' + a\(n A,<>f'') +fl2(Aier") = (rf + ajri -fa2)/t|er''

But this is equal to zero, and therefore satisfies equation (23,3). because the expression in brackets is the characteristic equation and n is chosen to set this expression equal to zero. Therefore we are certain that yi is a solution to equation (23.3),

We have found there are two distinct solutions to the homogeneous form of a linear, second-order differentia) equation with constant coefficients. However, we are trying to find one general solution, namely a solution that represents every possible solution. It turns out that we actually have found it. Theorem 23.1 provides the explanation.

Theorem 23.1 Lei yi and yj be two distinct .solutions to the differential equation (23.3). II C| and t'2 are any two constants, then ihe function y - o.Vi + qvv is a solution to equation (23.3). Conversely, if v is any solution to equation (23.3), then there are unique constants. Cj and t2, such thai y = ci yi + r2 y2,

Proof

We prove only the first part of Ihe theorem. If Vj and y2 are solutions to equation (23.3). then it follows that y> + «|.V| -I- «¡yi = y2 + «i.v2 + o2y2 = 0

We are given that y = C| V| + c2y2. If y is a solution, then it must be true that y + fl|.v + «2y = 0

Let us see if this is true. Use the definition of y to obtain y and y. then substitute to see if this equation is satisfied. Differentiating the definition of y gives v = fiy, -f c2y2

iuid differentiating again gives y = L'I_v | -f c'3y2

Substituting gives y 4- aiy + a2y = (C|>'| + c2y2) + o lù'iji'i + tyy2) + «2<fi.Vi + c2y2) = (y, -t- fl|>'| + a2y\ ) + C2('y2 + ai>'2 + «2.V2) = 0

Therefore y is a solution to equation (23.3). The second part of the theorem says thai any solution 10 the differential equation (e.g.. one satisfying specific initial conditions) can be expressed as a linear combination ofyi and y; through a suitable choice of the constants o and c2. Doing this requires that y, and y2 be distinct, by which we mean they must be linearly independent. ■

The implication of theorem 23.1 is that the general solution to the homogeneous form in equation (23.3) is yh = c,er{l + (23.10)

where we have defined new constants: C\ = f| Ai and C2 = c2/42. It is apparent now that we actually need two distinct solutions, yi and v2, in order to form the general solution. A rationale for this is that because two constants are lost in going from a function v(r ) to its second derivative, we actually need two distinct solutions so that these two constants can be recovered in solving the second-order differential equation,

IF r( - r2 (the case of repeated roots that occurs if — 4a2 = 0), wc do not have two distinct solutions, so theorem 23.1 would not seem to provide the general solution. However, it is still possible to find two distinct solutions. Rather than derive a second distinct solution, wc will simply state the result and then verify that it is correct.

If n = r2 = /•. the two distinct solutions to equation (23.3) are given by yi = A,erl and y2 = tA2er' (23.11)

These solutions are distinct because they are linearly independent. (Aie" cannot be made equal to tA2er' by multiplying it by a constant coefficient.) It is also possible to verify that the second solution will satisfy equation (23.3). To see this, first note that the ease of repeated roots arises only when

which means the solution to the characteristic equation is rt = r2 = r = -uJ2.

Next differentiate y2 to get y2 = A2e" + rt A2en

Differentiate again to get y2 = rA2e" + r A2en + r2tA2e"

Substitute y2 and its two derivatives (and the value of r) into equation (23.3) and check. The left-hand side of equation (23.3) becomes y+a\y + a2y = 2rA2en + r2tA2en + a\(A2er' + nA2e") + a2(rtA2er') = A2er'\t(r2 +ci,r +a2) + 2r + ai] = A2e"[/(a\/4 — a\/2 + a2) — ct\ 4-«,]

The last expression is equal to zero because a2 — 4a2 = 0 in the case of repeated roots. This proves that y2 is a solution for equation (23.3).

The results we have obtained to this point are summarized in theorem 23.2.

Theorem 23.2 |
The solution to the homogeneous form of the linear, second-order differential |

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