Ordinary Differential Equations (MA102 Mathematics II)

Shyamashree Upadhyay

IIT Guwahati

Shyamashree Upadhyay ( IIT Guwahati) Ordinary Differential Equations 1 / 10 Undetermined coefﬁcients-Annihilator Approach

A linear n-th order DE with constant coefﬁcients can be written as

n n−1 anD y + an−1D y + ··· + a1Dy + a0y = g(x) (1)

k dky where D y = dxk , k = 0, 1,..., n. This equation (1) can also be written as L(y) = g(x) where L denotes the linear n-th order differential operator

n n−1 L = anD + an−1D + ··· + a1D + a0. (2)

Factoring Operators: When the ai, i = 0, 1,..., n are real constants, the linear differential operator L in (2) can be factored whenever the characteristic polynomial n n−1 anm + an−1m + ··· + a1m + a0 factors. In other words, if r1 is a root of the auxilliary equation n n−1 anm + an−1m + ··· + a1m + a0 = 0, then L = (D − r1)P(D), where the polynomial expression P(D) is a linear differential operator of order n − 1. For example, if we treat D as an algebraic quantity, then the operator D2 + 5D + 6 can be factores as (D + 2)(D + 3) or as (D + 3)(D + 2). Thus if a function y = f (x) possesses a second derivative, then (D2 + 5D + 6)y = (D + 2)(D + 3)y = (D + 3)(D + 2)y.

Shyamashree Upadhyay ( IIT Guwahati) Ordinary Differential Equations 2 / 10 Annihilator Operator

This illustartes a general property: Factors of a linear differential operator with constant coefficients commute. If L is a linear differential operator with constant coefficients and f is a sufficiently differentiable function such that L( f (x)) = 0, then L is said to be an annihilator of the function f . For example, a constant function y = k is annihilated by D since Dk = 0. The function y = x is annihilated by D2 since the first and second derivatives of x are 1 and 0, respectively. Similarly, D3 x2 = 0, and so on. In general, the differential operator Dn annihilates each of the functions

1, x, x2,..., xn−1.

2 n−1 It hence follows that a polynomial c0 + c1 x + c2 x + ··· + cn−1 x can be annihilated by ﬁnding an operator that annihilates the highest power of x appearing in the polynomial.

Shyamashree Upadhyay ( IIT Guwahati) Ordinary Differential Equations 3 / 10 Annihilator Operator contd ...

In general, the functions that are annihilated by a linear n-th order differential operator L are simply those functions that can be obtained from the general solution of the homogeneous DE L(y) = 0.

Therefore, it follows that the differential operator (D − α)n annihilates each of the functions

eαx, xeαx, x2eαx,..., xn−1eαx.

To see this, note that the auxilliary equation of the homogeneous DE (D − α)ny = 0 is (m − α)n = 0. Since α is a root of multiplicity n of the auxilliary equation, therefore the general solution is αx αx n−1 αx y = c1e + c2 xe + ··· + cn x e .

Shyamashree Upadhyay ( IIT Guwahati) Ordinary Differential Equations 4 / 10 Annihilator Operator contd ...

Similarly, one can observe that the differential operator [D2 − 2αD + (α2 + β2)]n annihilates each of the functions

eαxcos βx, xeαxcos βx, x2eαxcos βx,..., xn−1eαxcos βx,

eαx sin βx, xeαx sin βx, x2eαx sin βx,..., xn−1eαx sin βx. When α = 0 and n = 1, it can be easily seen that as a special case of the above, we have

(D2 + β2)cos βx = 0 and (D2 + β2)sin βx = 0.

We are oftern interested in annihilating the sum of two or more functions.

If L is a linear differential operator such that L(y1) = 0 and L(y2) = 0, then L will annihilate the linear combination c1y1(x) + c2y2(x).

Shyamashree Upadhyay ( IIT Guwahati) Ordinary Differential Equations 5 / 10 Annihilator Operator contd ...

Let us now suppose that L1 and L2 are linear differential operators with constant coefﬁcients such that L1 annihilates y1(x) and L2 annihilates y2(x) but L1(y2) , 0 and L2(y1) , 0. Then the product L1L2 of differential operators annihilates the sum c1y1(x) + c2y2(x). We can easily show this, using linearity and the fact that L1L2 = L2L1:

L1L2(y1 + y2) = L1L2(y1) + L1L2(y2)

= L2L1(y1) + L1L2(y2) = L2(L1(y1)) + L1(L2(y2)) = 0. For example, D2 annihilates 7 − x and D2 + 16 annihilates sin 4x. Therefore the product D2(D2 + 16) will annihilate the linear combination 7 − x + 6sin 4x.

Shyamashree Upadhyay ( IIT Guwahati) Ordinary Differential Equations 6 / 10 Undetermined Coefﬁcients

Suppose that L(y) = g(x) is a linear differential equation with constant coefﬁcients and that the function g(x) is a linear combination of functions of the form

k(constant), xm, xmeαx, xmeαxcos βx, and xmeαx sin βx, where m is a non-negative integer and α, β are real numbers. We now know that such a function g(x) can be annihilated by a differential operator L1, of lowest order, consisting of a n n 2 2 2 n product of the operators D , (D − α) , and (D − 2αD + α + β ) . Applying L1 to both sides of the equation L(y) = g(x) yields L1L(y) = L1(g(x)) = 0. By solving the homogeneous higher order equation L1L(y) = 0, we can discover the form of a particular solution yp for the original non homogeneous equation L(y) = g(x). We then substitute this assumed form into L(y) = g(x) to ﬁnd an explicit particular solution. This method for determining yp is called the method of undetermined coefﬁcients, by Annihilator approach.

Shyamashree Upadhyay ( IIT Guwahati) Ordinary Differential Equations 7 / 10 Example

Solve: y00 + 3y0 + 2y = 4x2. (∗) −x −2x It is easy to see that the complementary function for (∗) is yc = c1e + c2e . Now since g(x) = 4x2 is annihilated by D3, we see that D3(D2 + 3D + 2)y = 4D3 x2 is the same as D3(D2 + 3D + 2)y = 0 (∗∗) The general solution of this ﬁfth order homogeneous equation (∗∗) is

2 −x −2x y = c1 + c2 x + c3 x + c4e + c5e . (∗ ∗ ∗)

The terms underlined in the above equation constitute the complementary function yc of the original equation (∗). We can then argue that a particular solution yp of (∗) should also satisfy equation (∗∗). This means that the terms remaining (not underlined) in equation (∗ ∗ ∗) must be the basic form of a particular solution yp of (∗) :

2 yp = A + Bx + Cx (#) where, for convenience, we have replaced c1, c2, c3 by A, B, C respectively.

Shyamashree Upadhyay ( IIT Guwahati) Ordinary Differential Equations 8 / 10 Example contd ...

For (#) to be a particular solution of (∗), we must have

00 0 2 yp + 3yp + 2yp = 4x

That is, we must have

2C + 3B + 6Cx + 2A + 2Bx + 2Cx2 = 4x2

Equating coefﬁcients of like powers of x, we get A = 7, B = −6, C = 2. Thus 2 yp = 7 − 6x + 2x . Hence the general solution of (∗) is y = yc + yp or

−x −2x 2 y = c1e + c2e + 7 − 6x + 2x .

Another problem: Solve y00 − 3y0 = 8e3x + 4sin x. 3x Clearly here yc = c1 + c2e .

Shyamashree Upadhyay ( IIT Guwahati) Ordinary Differential Equations 9 / 10 Example contd ...

Now since (D − 3)e3x = 0 and (D2 + 1)sin x = 0, we apply the product differential operator (D − 3)(D2 + 1) on both sides of the original equation given in the problem. That gives

(D − 3)(D2 + 1)(D2 − 3D)y = 0

This is a higher order homogeneous DE whose general solution is

3x 3x y = c1 + c2e + c3 xe + c4cos x + c5 sin x.

After excluding the linear combination of the underlines terms in the above equation (which correspond to yc), we arrive at the form of yp:

3x yp = Axe + Bcos x + Csin x.

Shyamashree Upadhyay ( IIT Guwahati) Ordinary Differential Equations 10 / 10 Table for Annihilators

A list of annihilators corresponding to various forms of g(x) g(x) Annihilator

n k n+1 Σk=0ak x D

n−1 k αx n Σk=0 ck x e (D − α)

n−1 αx k Σk=0 Ake x cos(βx)+ n−1 αx k 2 2 2 n Σk=0 Bke x sin(βx)[D − 2αD + (α + β )]

Here the ai (for i = 0, 1,..., n), ci (i = 0, 1,..., n − 1), Ak (k = 0, 1,..., n − 1), Bk (k = 0, 1,..., n − 1) are arbitrary real constants.

Shyamashree Upadhyay ( IIT Guwahati) Ordinary Differential Equations 11 / 10