Matrix Differential Operator Method of Finding a Particular Solution to A

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Although the problem of solving ordinary nonhomogeneous linear differential equations with constant coefficients is generally known, we will deal with the method of finding a particular solution of differential equation using a matrix differential operator method. This method, like the method of undetermined coefficients, can be used only in special cases, if the right-hand side of the differential equation is typical, i.e. it is a constant, a polynomial function, exponential function eαx, sine or cosine functions sin βx or cos βx, or finite sums and products of these functions (α, β constants). In some cases, the method can be used as a support method for determining the particular solution of a differential equation using the method of undetermined coefficients or the differential operator method or for evaluation of indefinite integrals. We introduce the differential operator notation. It is sometimes convenient to adopt the notation Dy, D2y, D3y, · · · , Dny dy d2y d3y dny to denote , , , · · · , . The symbols Dy, D2y,... are called differ- dx dx2 dx3 dxn ential operators [1] and have properties analogous to those of algebraic quan- tities [3].

Table 1: Operator Techniques [1], [10] A. n n φ(D) ckRk(x) ckφ(D)Rk(x) k=0 k=0 B. P P φ(D) · ψ(D) ψ(D) · φ(D) φ(D), ψ(D) polynomial operators in D C. φ(D)(xf(x)) xφ(D)f(x)+ φ′(D)f(x) D. Dneαx αneαx, n is non-negative integer E. Dn [eaxR(x)] eax (D + α)n R(x) n is non-negative integer F. n D2n sin βx −β2 sin βx n D2n cos βx −β2 cos βx Using the operator notation, we shall agree to write the diﬀerential equation

(n) (n−1) ′ any + an−1y + . . . + a1y + a0y = f(x), an 6= 0, ai ∈ R (1.1) as n n−1 anD + an−1D + · · · + a1D + a0 y = f(x) (1.2) MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 3 or brieﬂy φ(D)y = f(x), (1.3) where n n−1 φ(D)= anD + an−1D + · · · + a1D + a0 (1.4) is called an operator polynomial in D. If we want to emphasize the degree of the polynomial operator (1.4), we shall write it in the form φn(D).

The use of the operator calculus for solving linear diﬀerential equations is well dealt with in several publications, e.g. [3], [1], [10].

2. Matrix differential operator

Definition 1. If S = {v1, v2,..., vn} is a set of vectors in a vector space V, then the set of all linear combination of v1, v2,..., vn is called the span of v1, v2,..., vn and is denoted by span(v1, v2,..., vn) or span(S). Let G be a vector space of all differentiable function. Consider the subspace V ⊂ G given by V = span (f1(x),f2(x), · · · ,fn(x)) , (2.1) where we assume that the functions f1(x),f2(x), · · · ,fn(x) are linearly independent. Since the set B = {f1(x),f2(x), · · · ,fn(x)} is linearly independent, it is a basis for V. The functions fi(x), i = 1, 2,...,n expressed in basis B using base vector coordinates are usually written 1 0 0 0 1 0       [f1(x)]B = . , [f2(x)]B = . , · · · , [fn(x)]B = . . . .       0 0 1             The vector [fi(x)]B has in the i-th row 1 and 0 otherwise. Further, assume that the differential operator D maps V into itself Let n D(fi(x)) = cijfj(x), i = 1, 2, · · · , n, j X=1 where cij ∈ R, i, j = 1, 2,...,n are constants. Then

ci1 c  i2  [D (fi(x))]B = . , i, j = 1, 2,...,n .   cin     4 JOZEF FECENKO and (see [9])

c11 c21 · · · cn1 c c · · · c . . .  12 22 n2 [D]B = [[D (f1(x))]B . [D (f2(x))]B . · · · . [D (fn(x))]B]= ......   c1n c2n · · · cnn    (2.2) If n f(x)= αifi(x), αi ∈ R, i = 1, 2, · · · ,n, f(x) ∈ V i X=1 then α1 α  2 [f(x)]B = . .   αn   We express what is the derivative of the function  f(x) n n n n n Df(x)= αiDfi(x)= αi cijfj(x)= αicijfj(x) i i j j i X=1 X=1 X=1 X=1 X=1 respectively n ci1αi i=1 n c11 c21 · · · cn1 α1  P  ci2αi c12 c22 · · · cn2 α2 [D(f(x))] = i=1  =  . . .   .  B  .  ......  P .  . . . . .  .       n  c1n c2n · · · cnn αn        cinαi     i=1  P  Let us next simply denote [D]B as DB. The matrix DB we will called a matrix diﬀerential operator corresponding to a vector space V with the considered basis B. Denote β1 β ′  2  [D(f(x)]B = f B = . and [f(x)]B = f B .   βn   then   ′ f B = DBf B MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 5

Note that the matrix transformation DB : V → V deﬁned by

′ DBf B = f B is a linear transformation. As mentioned in (2.2), i-th, (i = 1, 2, · · · ,n) column of the matrix D expresses the derivative of the function fi(x). The following properties of the matrix diﬀerential operator are given with- out proof.

Theorem 1. Let DB be a matrix diﬀerential operator of vector space V with a basis B. Then

• DB (f B + gB)= DBf B + DBgB R • DB (cf B)= cDB (f B) , c ∈ ′′ 2 • f B = DB (DBf B)= DBf B (n) (n−1) (0) • f B = DBf B , f B = f B (n) n • f B = DBf B n • i-th (i = 1, 2, · · · ,n) column of the matrix DB expresses n-th derivative of function fi(x).

Example 1. Let us consider

V = span(xe2x sin 3x,xe2x cos 3x, e2x sin 3x, e2x cos 3x) (2.3)

The Wronskian W (xe2x sin 3x,xe2x cos 3x, e2x sin 3x, e2x cos 3x) = 324 · e8x 6= 0. It follows that the functions in (2.3) are linearly independent. Then the set B = {xe2x sin 3x,xe2x cos 3x, e2x sin 3x, e2x cos 3x} is a basis for V. Applying diﬀerential operator D to a general element of V, we see that

D axe2x sin 3x + bxe2x cos 3x + ce2x sin 3x + de2x cos 3x 2x 2x 2x = (2a − 3b)xe sin 3x + (3a + 2b)xe cos 3x + (a + 2c − 3d)e sin 3x + (b + 3c + 2d)e2x cos 3x, which is again in V. We found the corresponding matrix diﬀerential operator DB in the vector space V with the basis B. Here the construction of DB is shown schematically 6 JOZEF FECENKO x x x x cos3 sin 3 cos3 sin 3 x x 2 2 x x 2 2 xe e e xe D D D D xe2x sin 3x 2 −3 0 0 xe2x cos 3x 3 2 0 0 D = (2.4) B e2x sin 3x 1 0 2 −3 e2x cos 3x 0 1 3 2    Let us calculate the ﬁrst and second derivative of the functi on

f(x)= xe2x sin 3x + 2xe2x cos 3x − 2e2x cos 3x using the matrix diﬀerential operator (2.4). We have

2 −3 0 0 1 −4 3 2 0 0 2 7 [Df(x)] = D [f(x)] = = B B B 1 0 2 −3  0   7  0 1 3 2  −2 −2             So that

Df(x)= −4xe2x sin 3x + 7xe2x cos 3x + 7e2x sin 3x − 2e2x cos 3x

The second derivative

2 −3 0 0 −4 −29 ′′ ′ 3 2 0 0 7 2 f = D f = = B B B 1 0 2 −3  7   16  0 1 3 2  −2  24             

D2f(x)= −29xe2x sin 3x + 2xe2x cos 3x + 16e2x sin 3x + 24e2x cos 3x

For calculating higher powers of the matrix DB it is sometimes advanta- geous to express it as a block matrix (see example 11) or use Jordan matrix decomposition. The point of a matrix differential operator is not that this method is easier than a direct differentiation. Indeed, once the matrix DB has been established, it is easy to find the differentials with little to do. What is significant is that matrix methods can be used at all in what appears, on the surface, to be a calculus problem. MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 7

3. Some properties of block matrices and pseudoinverse matrix Theorem 2. [5] Let A be a matrix partitioned into four 2 × 2 blocks

P Q A = R S where R is a regular matrix of the order r. Then the rank of matrix A is equal to r if and only if − Q = P R 1S. Proof. To the first block row of the matrix A we add the second block row multiplied from the left by the matrix −P R−1. This adjustment adds the linear combination of matrix rows (R S) to the other matrix rows. After this adjustment, we get the matrix 0 Q − P R−1S . (3.1) R S For matrix A and matrix (3.1) to have the same rank, the matrix Q − P R−1S must be equals 0. It follows that Q = P R−1S. Definition 2. Let A be a m × n matrix. The matrix X for which AXA = A, is called the pseudoinverse matrix of matrix A. The pseudoinverse matrix of a matrix A is denoted A−. Theorem 3. Let A be a m × n matrix partitioned into four 2 × 2 blocks P Q A = R S and let the rank of the matrix A be r. If the matrix R is regular of order r, then the n × m matrix −1 − 0 R A = 0 0 is the pseudoinverse of the matrix A. Proof. Compute −1 −1 − P Q 0 R P Q P P R S P Q AA A = = = = A R S 0 0 R S R S R S We have used the consequence of the Theorem 2, i.e. Q = P R−1S. − Definition 3. The pseudoinverse matrix A of the matrix A satisfying all of the following four conditions (i) AA−A = A, (A− is the pseudoinverse matrix of the matrix A); 8 JOZEF FECENKO

(ii) A−AA− = A−, (A is the pseudoinverse matrix of the matrix A−); T (iii) AA− = AA−, (AA− is a symmetric matrix, (·)T means a trans- posed matrix); T (iv) A−A = A−A, (A−A is a symmetric matrix) + is called the Moore-Penrose pseudoinverse of the matrix A, denoted by A . Theorem 4. Let A be a m × n matrix partitioned into four 2 × 2 blocks 0 0 A = R 0 and let the rank of matrix A be r. If matrix R is regular of order r, then −1 − 0 R A = 0 0 n × m is the Moore-Penrose pseudoinverse matrix of the matrix A, thus A− = A+. Proof. We will verify all four conditions of deﬁnition 2. 0 0 0 R−1 0 0 0 0 (i) AA−A = = = A R 0 0 0 R 0 R 0 The condition i is met. (ii) Pseudoinverse matrix of the matrix −1 − 0 R A = 0 0 is the matrix 0 0 A = , R 0 because −1 −1 −1 − − 0 R 0 0 0 R 0 R − A AA = = = A 0 0 R 0 0 0 0 0 We have also proved the validity of condition ii. −1 T T T 0 0 0 R 0 0 0 0 (iii) AA− = = = R 0 0 0 0 I 0 I We have shown that condition iii is met. −1 T T T 0 R 0 0 I 0 I 0 (iv) A−A = = = 0 0 R 0 0 0 0 0 We have shown that condition iv is also met. Corollary 1. For the matrices A and A− in Theorem 4 imply that AA− and A−A are incomplete identity matrices. MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 9

Theorem 5. Let Ax = b (3.2) be a solvable system of linear equations and A− be a pseudoinverse of the matrix A, then − x = A b (3.3) is a solution of (3.2). Proof. Assuming that the system of linear equations (3.2) is solvable, then a vector y exists, such that Ay = b We show that the vector in (3.3) is the solution of the system of linear equations (3.2). It is valid that − − Ax = A A b = AA A y = Ay = b Remark 1. If the system of linear equations (3.2) is solvable and its solution is − − expressed in the form x = A b and A is Moore-Penrose pseudoinverse of the matrix A, then this solution minimizes the Euclidean norm kAx − bk and of all n-dimensional solutions x that minimize this norm it has the lowest norm. Such a vector solution of the system of linear equations is called the solution of the system of linear equations with the least squares solution of minimum norm or the best approximate solution of the system. Note that in our considerations, when solving a solvable system of linear equations (3.2) we will consider a solution (3.3) expressed in the form x = A+b The solution of (3.2) exists if and only if AA+b = b, [4].

4. Pseudoinverse matrix differential operator We shall introduce some properties of the differential and inverse differential operators. 1 The definition of the inverse differential operator [10]: Let f(x) be φ(D) 1 defined as a particular solution y, such that (1.3) then f(x) is called the φ(D) inverse differential operator. We will prove relation I in the next Table 2. 10 JOZEF FECENKO

Proof. With respect to C in the Table 1 is

′ φ(D)(xf(x)) = xφ(D)f(x)+ φ (D)f(x) (4.1)

Let

φ(D)f(x)= R(x)

Then

1 1 φ(D)f(x)= R(x) φ(D) φ(D)

1 f(x)= R(x) (4.2) φ(D)

Substituting (4.2) into relationship (4.1), we get

1 1 ′ 1 φ(D) x R(x) = xφ(D) R(x)+ φ (D) R(x) φ(D) φ(D) φ(D) 1 ′ 1 φ(D) x R(x) = xR(x)+ φ (D) R(x) φ(D) φ(D)

1 Let’s multiply the previous equation from the left side with , then φ(D)

1 1 1 ′ 1 x R(x)= xR(x)+ φ (D) R(x) φ(D) φ(D) φ(D) φ(D)

From which

1 1 1 ′ 1 xR(x)= x R(x) − φ (D) R(x) φ(D) φ(D) φ(D) φ(D)

MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 11

Table 2: Inverse Operator Techniques [1], [10] A. n n 1 1 φ(D) ckRk(x) ck φ(D) Rk(x) k=0 k=0 B. P P 1 mx −mx D−m R(x) e e R(x)dx R C. − − em1x e m1x em2x e m2x . . . 1 mnx −mnx n − − − e e R(x)dx (D m1)(D m2)...(D mn) R R This can also be evaluated by expanding R the inverse operator into partial fractions and then using B. D. epx φ(p) if φ(p) 6= 0 k px ′ − 1 epx x e if φ(p)= φ (p)= ...φ(k 1)(p) = 0 φ(D) φ(k)(p) but φ(k)(p) 6= 0 E. cos px 1 cos px φ(D2) φ(−p2) if φ(−p2) 6= 0 sin px 1 sin px φ(D2) φ(−p2) F. 1 eipx φ(D) cos px Re φ(ip) 1 n eipx o φ(D) sin px Im φ(ip) If φ(ip) 6=n 0, otherwiseo use D. G. p 1 p k p φ(D) x ckD x k=0 1 p+n p by expanding φ(D) in powers of D since DP x = 0 for n> 0. H. 1 px px 1 φ(D) e R(x) e φ(D+p) R(x) called the ”operator shift theorem”. I. 1 1 1 ′ 1 φ(D) xR(x) x φ(D) R(x) − φ(D) φ (D) φ(D) R(x)= 1 ′ 1 x − φ(D) φ (D) φ(D) R(x) J. using substitute 1 eiβx + e−iβx (eαxP (x) cos βx+ cos βx = φ(D) m 2 eiβx − e−iβx Q (x)sin βx) sin βx = n 2i and next to use an operator shift theorem 12 JOZEF FECENKO

Proofs of many of the statements in Table 2 can be found for example in [1] or [10]. Note that all analytical relations derived for the differential operator have an adequate expression for the matrix differential operator. In order to define a pseudoinverse matrix differential operator, we first consider a simple form of the differential equation (1.1) i. e. ′ y = f(x) (4.3) Let us assume that the function f(x) is differentiable and every derivative of the function f(x) can be expressed as a finite linear combination of linear independent functions f1(x),f2(x),...,fn(x). Let’s consider the vector space

V = span (f1(x),f2(x),...,fn(x)) with the basis B = {f1(x),f2(x),...,fn(x)}. (A) Let us assume that the particular solution of the diﬀerential equation (4.3) belongs to the vector space V with the basis B. Let DB be a matrix diﬀerential operator corresponding to the basis B. Then the equation

DByB = f B,

where f B = [f(x)]B, is solvable. And due to the Theorem 5 + yB = DB f B (4.4) is the solution of the differential equation (4.3) expressed in the basis + B where DB is the Moore-Penrose pseudoinverse of the matrix DB. + Thus DB is the pseudoinverse matrix differential operator to the operator DB . (The unique solution (4.4) to the differential equation (4.3) + do not contain a kernel of matrix differential operator DB. This also applies in the following cases.) (B) Let us assume that the particular solution of the differential equation (4.3) does not belong to the vector space V and let DB be a matrix differential operator with the considered basis B. Then the equation

DByB = f B is not solvable. Let’s create a new system of functions

B1 = {f1(x),f2(x),...,fn(x)} ∪ {xf1(x),xf2(x),...,xfn(x)}

= {f11(x),f12(x),...,f1m(x)}

where m> n. Let functions f11(x),f12(x),...,f1m(x) be a linear independent and let

V1 = span(f11(x),f12(x),...,f1m(x)) MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 13

with the basis B1. Let a particular solution of the diﬀerential equation

(4.3) belong to the vector space V1 and let DB1 be a matrix differential operator with the considered basis B1. Then the equation y f DB1 B1 = B1 (4.5) f is solvable, where B1 = [f(x)]B1 . Due to Theorem 5 y = D+ f B1 B1 B1 is the solution of the differential equation (4.3) expressed in the basis B where D+ is the Moore-Penrose pseudoinverse of the matrix 1 B1 + DB1 . Thus DB is the pseudoinverse matrix differential operator to the operator DB . (C) If a solution of the differential equation (4.3) does not belong to the vector space V1, we will create a new system of vectors in a manner similar to B1 and analyse the solvability of the differential equation. It is difficult to prove in general that the solution lies in some vector space, the construction of which we have described. It will be explained in the following examples. For completeness, we still have to consider a differential equation in the form (1.1). Also in this case, we assume that the function f(x) is differentiable and every derivative of the function f(x) can be expressed as a finite linear combination of linear independent functions f1(x),f2(x),...,fn(x). We are considering a vector space

V = span(f1(x),f2(x),...,fn(x)) with the basis B = {f1(x),f2(x),...,fn(x)} and and we discuss the existence of a solution to the equation n n−1 anDB + an−1DB + · · · + a1DB + a0I yB = f B (4.6) or brieﬂy

φ(DB)yB = f B, where I is the identity matrix. The discussion is analogous to previous cases. Now let us present some elementary examples. We will focus this issue in more detail below. Let’s start with a very elementary example. Example 2. Determine using a matrix diﬀerential operator the particular solution of the equation ′ y = x. (4.7) 14 JOZEF FECENKO

Solution. f(x) = x and each derivative of f(x) can be expressed as a linear combination of {x, 1}. Let V = span(x, 1). Then 0 0 D = . B 1 0 The equation DByB = [x]B 0 0 y 1 1 = 1 0 y 0 2 has no solution. It follows that a particular solution of (4.7) does not belong 2 2 to V . Let’s create a new system of function B1 = {x ,x} ∪ {x, 1} = {x , x, 1} which is linear independent. Then 0 0 0 2 V1 = span(x , x, 1), DB1 = 2 0 0 0 1 0. The matrix equation   y DB1 B1 = [x]B1 0 0 0 y1 0 2 0 0 y = 1    2   0 1 0 y3 0 has a solution and (Theorem 5)     

y1 0 + y = y2 = D 1 B1   B1   y3 0 According to Theorem 4     1 0 0 D+ = 2 B1  0 0 1  0 0 0   Thus   1 1 0 0 0 y = 2 1 = 2 B1 0 0 1   0 0 0 0 0 0     which, in turn, implies the particular solution     1 y = x2 2 Example 3. Determine the particular solutions of the equations (a) y′′ + 3y′ − 4y = xe2x MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 15

(b) y′′ + 3y′ − 4y = 2xe2x − 3e2x Solution. Every derivative of the functions on the right-hand side of the equations is a linear combination of the functions x · e2x, e2x which are linear independent Let’s assume that a particular solution of the diﬀerential equations belongs to the vector space V = span(xe2x, e2x) with the base B = {xe2x, e2x}. Diﬀerential operator 2 0 D = B 1 2 (a) 2 2x DB + 3DB − 4I2 yB = [xe ]B

2 0 2 2 0 1 0 1 + 3 − 4 y = 1 2 1 2 0 1 B 0 ! 6 0 1 y = 5 −4 B 0

− 1 1 6 0 1 1 0 1 y = =  6    =  6  B 5 −4 0 5 1 5  −  0   24 4   24 We have found the particular solution of diﬀerential   equatio n in (a)

1 5 y = xe2x + e2x. 6 24 (b) 1 1 0 2 y =  6    = 3 B 5 1 7  −  −3   24 4   6  1   7    y = xe2x + e2x 3 6 Example 4. Find the particular solution of the diﬀerential equation ′′ yIV + 2y + y = 2sin x − 4 cos x (4.8) Solution. It’s easy to prove that the particular solution y of diﬀerential equation (4.8) y∈ / V = span(sin x, cos x) y∈ / V = span(x sin x,x cos x, sin x, cos x) 16 JOZEF FECENKO

Let’s assume that y ∈ V = span(x2 sin x,x2 cos x,x sin x,x cos x, sin x, cos x). Since the set of functions

B = {x2 sin x,x2 cos x,x sin x,x cos x, sin x, cos x} is linearly independent, it is a basis for V . The corresponding matrix diﬀer- ential operator 0 −10 0 0 0 100000 2 0 0 −1 0 0  D = B 021000    00100 −1   000110      Then

4 2 DB + 2 · DB + I6 yB = [2 sin x − 4 cos x]B 0 0 0000 0 0 0 0000 0  0 0 0000  0  y =  0 0 0000 B  0      −8 0 0000  2       0 −80000 −4     +  0 0 00 00 0  0 0 00 00 0     0 0 00 00 0 y = B  0 0 00 00   0       −8 0 00 00   2       0 −8 00 00  −4      1    1 00 00 − 8 0 0 − 4 00 00 0 − 1 0 1  8     2  00 00 0 0 0 0 y = = B  00 00 0 0   0   0         00 00 0 0   2   0         00 00 0 0  −4  0              4 2 The Moore-Penrose pseudoinverse to the matrix DB + 2 · DB + I6 we found using Theorem 4. Then 1 1 y = − x2 sin x + x2 cos x 4 2 MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 17

5. Matrix differential operator and the method of undetermined coeficients In order to describe the algorithm for the finding of a particular solution of an ordinary nonhomogeneous linear differential equation with constant coefficients (1.2) using matrix differential operator, we use knowledge of the algorithm for finding a particular solution by an undetermined coefficients method. Let us consider a differential equation (1.2) with a characteristic equation n n−1 ank + an−1k + . . . + a1k + a0 = 0 (5.1) First, we make two conventions to simplify expressing the root multiplicity of the characteristic equation and the value of the function f(x) = x0 at the discontinuity point. 1. If the number α is not the root of the characteristic equation (5.1), we will also say that it is a 0-fold root of the characteristic equation; if the number α is a simple characteristic root of (5.1), we will also say that it is 1-fold root of the characteristic equation, and so on. 2. Given the removable discontinuity of the function f(x) = x0 at the point x = 0, we define the function at this point as f(0) = 1. Let us consider two cases of the right-hand side of the differential equation (1.2). We will describe the algorithm for finding a particular solution using a matrix differential operator: a) with the right-hand side (1.2) αx f(x)= e Pm(x) (5.2)

where α ∈ R and Pm(x) is a polynomial of degree m. The algorithm to determine the particular solution of that differential equation, by the method of undetermined coefficients, says: If α is the k-fold root (k = 0, 1, 2,...,) of the characteristic equation (5.1) of the differential equation (1.2) with the right-hand side (5.2), then the particular solution of this equation is of the form k αx y = x e Qm(x), (5.3) m m−1 where Qm(x) = Amx + Am−1x + . . . + A1x + A0 is a polynomial of degree m with undetermined coefficients. The values of these undetermined coefficients can be determined by substituting (5.3) for y into the equation (1.2) and then comparing the coefficients for the same functions on the right-hand and left-hand sides of the equation. The algorithm for determining the particular solution of the differential equation (1.2) with the right-hand side (5.3) by the method of 18 JOZEF FECENKO

undetermined coeﬃcients implies that this particular solution will be in the vector space − V = span(xk+meαx,xk+m 1eαx,...,xeαx, eαx) with the basis for V − B = xk+meαx,xk+m 1eαx,...,xeαx, eαx (5.4) n o The relevant matrix diﬀerential operator DB in the vector space V with the basis B be a matrix of the type (k + m + 1) × (k + m + 1). b) with the right-hand side αx f(x)= e (Pr(x)sin βx + Qs(x) cos βx) , (5.5)

where α, β are real numbers, Pr(x) is a polynomial of degree r, Qs(x) is a polynomial of degree s. If α+βi is a k-fold root (k = 0, 1, 2,...,) of the characteristic equation (5.1) of the differential equation (1.2) with the right-hand side (5.5), then the particular solution of this equation has the form k αx y = x e (Um(x)sin βx + Vm(x) cos βx) (5.6) m m−1 where m = max {r, s}, Um(x)= Amx + Am−1x + . . . + A1x + A0,, m m−1 Vm(x)= Bmx + Bm−1 x + . . . + B1x + B0 are polynomials of degree m with undetermined coefficients. The values of these undetermined coefficients can be determined by substituting (5.6) for y into the equation (1.2) and then comparing the coefficients for the same functions on the right-hand and left-hand sides of the equation. The algorithm for determining the particular solution of the differential equation (1.2) with the right-hand side (5.6) by the method of undetermined coefficients implies that this particular solution will be in the vector space V = span(xk+meαx sin βx,xk+meαx cos βx,xk+m−1eαx sin βx, xk+m−1eαx cos βx...,xeαx sin βx,xeαx cos βx,eαx sin βx, eαx cos βx) with the basis for V B = xk+meαx sin βx,xk+meαx cos βx,xk+m−1eαx sin βx, xk+m−1eαx cos βx...,xeαx sin βx,xeαx cos βx,eαx sin βx, (5.7) eαx cos βx}

The relevant matrix differential operator DB in the vector space V with the basis B be a matrix of the type 2(k + m + 1) × 2(k + m + 1). Subsequently, in both cases a) and b), we create a matrix equation n n−1 anDB + an−1DB + · · · + a1DB + a0I yB = f B (5.8) MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 19 where f B = [f(x)]B and yB = [y(x)]B, where y(x) is a particular solution of the differential equation (1.2). If the right-side of the differential equation (1.1) is the sum of several functions, then the principle of superposition can be used to solve it. The principle of superposition of solutions says that if yi (i = 1, 2,...m) is a solution of the differential equation

(n) (n−1) ′ any + an−1y + · · · + a1y + a0 y = fi(x) (i = 1, 2,...m), then for any constants k1, k2,...,km, the function

y = k1y1 + k2y2 + · · · + kmym is a solution to the diﬀerential equation (1.1) with

f(x)= k1f1(x)+ k2f2(x)+ · · · + kmfm

In special case, if f1(x),f2(x),...,fm(x) form the basis B of the vector space V = span(f1(x),f2(x),...,fm(x)) then it is enough to solve only the equation

k1 k n n−1  2  anDB + an−1DB + · · · + a1DB + a0I yB = . .   km  B   In the next example, we want to show how the matrix differential operator can be used to support the finding of a particular solution of the differential equation by the method of undetermined coefficients.

Example 5. Determine the particular solution of the diﬀerential equation

(D − 2)2(D + 4)2y = 3e2x (5.9)

Solution. First we solve the equation

(D + 4)2y = 3e2x (5.10)

Since α = 2 is not a solution of the characteristic equation (k + 4)2 = 0, it follows that the solution of (5.10) belongs to the vector space V = span(e2x) 2x with the basis B1 = {e }. Then the matrix diﬀerential operator is

DB1 = [2] 20 JOZEF FECENKO and 2 2x I y1 (DB1 + 4 1) B1 = [3e ]B1 2 y1 ([2] + 4[1]) B1 = [3] 2 1 [6] y B1 = [3] −2 y1 B1 = [6] [3] 1 y1 = B1 12 1 y = e2x 1 12 Now we have to determine the particular solution of the differential equation 1 (D − 2)2y = e2x (5.11) 12 Because α = 2 is 2-fold root of the characteristic equation (k − 2)2 = 0, it follows that the solution of (5.11) belongs to the vector space V = span(x2e2x, xe2x, e2x) with the basis B = {x2e2x, xe2x, e2x}. Then the matrix differential operator is 2 0 0 B = 2 2 0 0 1 2 and   2 1 2x (DB − 2I ) y = e 3 B 12 B 0 0 0 0 0 0 0 y = 0   B  1  2 0 0 12     0 0 0 + 0 y = 0 0 0 0 B  1   2 0 0  12     1 1 0 0 0 y = 2 0 = 24 B  0 0 0   1   0  0 0 0 0   12   The particular solution of the differential  equation  (5.9) is 1 y = x2e2x 24 MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 21

Example 6. Determine the particular solution of the differential equation ′′ ′ y − 4y + 13y = 2xe2x cos 3x (5.12) using a pseudoinverse matrix differential operator. Solution. Because 2 + 3i is 1-fold root of the characteristic equation k2 − 4k + 13 = 0, it follows that the particular solution of (5.12) will be (due to (5.5),(5.6)) in the form y = xe2x ((Ax + B)sin βx + (Cx + D) cos βx) In other words, the particular solution y will be in the vector space V = span(x2e2x sin 3x,x2e2x cos 3x,xe2x sin 3x,xe2x cos 3x, e2x sin 3x, e2x cos 3x) with the basis B = x2e2x sin 3x,x2e2x cos 3x,xe2x sin 3x,xe2x cos 3x, e2x sin 3x, e2x cos 3x The matrix differential operator 2 −30 0 0 0 320000   2 0 2 −3 0 0 D = B 023200    00102 −3   000132      2 2x DB − 4DB + 13I 6 yB = [2xe cos 3x]B After editing the previous equation, we get 0 0 00 0 0 0 0 0 00 0 0 0     0 −12 0 0 0 0 0 y =  12 0 0 0 0 0  B 2      2 0 0 −6 0 0  0      0 2 60 0 0  0         + 0 0 00 0 0 0 0 0 00 0 0 0     0 −12 0 0 0 0 0 y = B  12 0 0 0 0 0  2      2 0 0 −6 0 0  0      0 2 60 0 0  0         22 JOZEF FECENKO

In accordance with Theorem 4 we determine the Moore-Penrose pseudoinverse. Then 1 1 0 0 0 0 0 0 12 6  1      0 0 − 0 0 0 0 0  12             1 1       0 0 0 0  0  0  y =  36 6    =   B  1 1     1   0 0 0 − 0  2          36 6    18        0 0 0 0 00  0  0                     0 0 0 0 00  0  0              So, the particular solution is

1 2 2x 1 2x yp = x e sin 3x + xe cos 3x 6 18 Example 7. Find the integral

13xe2x sin 3x − 13xe2x cos 3x + 5e2x sin 3x − 4e2x cos 3x dx Z Solution. We want to solve the diﬀerential equation Dy = 13xe2x sin 3x − 13xe2x cos 3x + 5e2x sin 3x − 4e2x cos 3x The solution will be in the vector space V = span(xe2x sin 3x,xe2x cos 3x,e2x sin 3x,e2x cos 3x) with the basis B = xe2x sin 3x,xe2x cos 3x, e2x sin 3x, e2x cos 3x In (2.4) we calculated 2 −3 0 0 3 2 0 0 D = B 1 0 2 −3 0 1 3 2    We are having to solve the matrix equation  13 −13 D y = B B  5   −4      MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 23

Then 2 3 0 0 13 −1 13 13 13  3 2      − 0 0 −13 −5 −1 −13  13 13      yB = DB   =     =   5  5 12 2 3     15   −   5     −4   169 169 13 13     13             12 5 3 2     16  −   −4  −   169 169 13 13     13       We have calculated that

13xe2x sin 3x − 13xe2x cos 3x + 5e2x sin 3x − 4e2x cos 3x dx Z 15 16 = −xe2x sin 3x − 5xe2x cos 3x + e2x sin 3x − e2x cos 3x + C 13 13 Sometimes it is useful to combine a differential operator with a matrix differential operator. Relationship I in Table 2 allows us to reduce a matrix differential operator by two rows and two columns. In the following we will give an example of this. Example 8. Find a particular solution of (D2 − 5D + 16)y = xe2x sin 3x (5.13) Solution. Using I in the Table 2 we have 1 y = xe2x sin 3x D2 − 5D + 16 (5.14) 1 1 = x − (2D − 5) e2x sin 3x D2 − 5D + 16 D2 − 5D + 16 All we need to do is to create a 2 × 2 matrix differential operator instead of 4 × 4. The particular solution of (5.13) belongs to the vector space V = span xe2x sin 3x,xe2x cos 3x, e2x sin 3x, e2x cos 3x . It can be reduced using (5.14) to the vector space V = span e2x sin 3x, e2x cos 3x with the basis B = {e2x sin 3x, e2x cos 3x}. Then 2 −3 D = B 3 2 1 3 D2 − 5D + 16I = B B 2 −3 1 24 JOZEF FECENKO

1 3 − 2 −1 D − 5DB + 16I2 = 10 10 B 3 1   10 10    − −1 −6 (2D − 5I ) 1 = B 2 6 −1 where I2 is the 2 × 2 identity matrix. The right-hand side of the relationship (5.14) expressed by the diﬀerential operator is

−1 −1 1 xI − D2 − 5D + 16I (2D − 5I D2 − 5D + 16I = 2 B B 2 B 2 B B 2 0 1 3 1 3 1 0 − −1 −6 − 1 x   − 10 10   10 10   = 3 1 3 1  0 1    6 −1   0    10 10    10 10               1 3 1 x 0 − −1 −6  − 10 10   10 = 0 x 3 1 3     6 −1    10 10    10        1 1 3 19 1 7 x − − x + 10  − 10 10  10 = 10 25 3 3 1 3 3 27  10 x      x +    10 10   10  10 50 We can rewrite this using functions  as    1 1 7 3 27 xe2x sin 3x = x + e2x sin 3x + x + e2x cos 3x D2 − 5D + 16 10 25 10 50 so the particular solution of the diﬀerential equation (5.13) is 1 7 3 27 y = x + e2x sin 3x + x + e2x cos 3x 10 25 10 50 In the previous example, the number resulting from the right-hand side of the diﬀerential equation (5.13) was not the root of the corresponding characteristic equation of (5.13). In the next example it will be. Example 9. Determine the particular solution of the equation D2 + 1 y = x cos x (5.15) MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 25

Solution. The complex number i is a single root of the characteristic equation k2 + 1 = 0, therefore the particular solution of the differential equation (5.15) will be in the vector space V = span x2 sin x,x2 cos x,x sin x,x cos x, sin x, cos x with the basis for V B = {x2 sin x,x2 cos x,x sin x,x cos x, sin x, cos x} The pseudoinverse matrix differential operator will be a 6 × 6 matrix. We will present three methods of solution: using a matrix differential operator, a combined method and using a complex variable a) using a matrix differential operator We have the matrix differential operator 0 −10 0 0 0 100000 2 0 0 −1 0 0  D = B 021000    00100 −1   000110      2 (DB + I6)yB = [x cos x]B 0 0 0 0 0 0 0 0 0 0 0 0 0 0     0 −4 0 0 0 0 0 y =  4 0 0 0 0 0  B 1      2 0 0 −2 0 0  0      0 2 2 0 0 0  0     Using a Moore-Penrose pseudoinverse matrix (Theorem  4 and 5) we have 2 + yB = (DB + I6) [x cos x]B

1 1 0 0 0 4 0 0 0 4 0 0 − 1 0 0 0 0 0  4      0 0 1 0 0 1 0 0 y = 4 2 = B  0 0 0 1 − 1 0  1  1   4 2     4   0 0 0 0 0 0  0 0        0 0 0 0 0 0  0 0       So the particular solution of (5.15) is      1 1 y = x2 sin x + x cos x 4 4 26 JOZEF FECENKO b) We can reduce the size of the matrix DB by using I in the Table 2. 1 1 1 1 x cos x = x cos x − 2D cos x D2 + 1 D2 + 1 D2 + 1 D2 + 1 1 Let’s solve cos x which corresponds to the particular solution of the D2 + 1 diﬀerential equation (D2 + 1)y = cos x. This particular solution belongs to the vector space V = span(x sin x,x cos x, sin x, cos x) with the basis B = {x sin x,x cos x, sin x, cos x} Then 0 −1 0 0 1 0 0 0 D = B 1 0 0 −1 0 1 1 0    2  DB + I4)yB = [cos x]B 0 0 0 0 0 0 0 0 0 0   y =   0 −2 0 0 B 0  2 0 0 0  1       +   0 0 0 0 0 0 0 0 0 0 y =     B 0 −2 0 0 0  2 0 0 0  1      1    1 0 0 0 2 0 2 0 0 − 1 0 0 0 y =  2    =   B 0 0 0 0 0 0  0 0 0 0  1 0       We have calculated that      1 x sin x cos x = D2 + 1 2 Similarly 1 0 0 0 2 0 0 1 0 0 − 1 0 0 − 1 sin x =  2    =  2  D2 + 1 0 0 0 0 1 0 B  0 0 0 0  0  0        So       1 x cos x sin x = − D2 + 1 x MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 27

Let’s continue to calculate 1 x sin x 1 1 x cos x = x − 2D cos x D2 + 1 2 D2 + 1 D2 + 1 1 x sin x 1 x cos x = x − (sin x + x cos x) D2 + 1 2 D2 + 1 1 x sin x x cos x 1 x cos x = x + − x cos x D2 + 1 2 2 D2 + 1 1 From this, by expressing x cos x, we have D2 + 1 1 x2 sin x x cos x x cos x = + D2 + 1 4 4 c) using a complex variable. To use J in the Table 2. See also [1]. 1 1 eix + e−ix y = x cos x = x D2 + 1 D2 + 1 2 1 1 − 1 = eix x + e ix x 2 (D + i)2 + 1 (D − i)2 + 1 1 1 − 1 = eix x + e ix x 2 D(D + 2i) D(D − 2i) 1 1 i D − 1 i D = eix − + x + e ix + x 2 D 2 4 D 2 4 1 1 ix 1 − 1 ix 1 = eix − + + e ix + 2 D 2 4 D 2 4 2 2 1 ix x − ix x = eix − + + e ix + 2 4 4 4 4 ix2 −eix + e−ix x eix + e−ix = + 4 2 4 2 x2 eix − e−ix x eix + e−ix x2 sin x x cos x = + = + 4 2i 4 2 4 4 In the previous example we compared the solution method with other methods. In the following example, we return to the diﬀerential operator for solving a linear ODE with a right-hand polynomial function. In this case, it is pro- 1 posed to expand the operator φ(D) in powers of D. For example, in [1] for 1 1 f(x)= n n−1 f(x) φ(D) anD + an−1D + . . . + a1D + a0 28 JOZEF FECENKO is the expand for a0 6= 0 ∞ 1 1 a a k y = f(x)= (−1)k n Dn + . . . + 1 D f(x) (5.16) φ(D) a0 a0 a0 Xk=0 In the case a0 = a1 = . . . = ak−1 = 0 (1 ≤ n ≤ n)) and (ak 6= 0, then

1 −k 1 y = f(x)= D n−k n−k−1 f(x), (5.17) φ(D) anD + an−1D + . . . + ak where we would expand an inverse operator as a Maclaurin series in the sense of (5.16). The disadvantage (5.16) of expanding as a Maclaurin series is the calculation a a k n Dn + . . . + 1 D although it is not necessary to count all members of a0 a0 the power. We will show a diﬀerent approach to solving this problem.

Theorem 6. Let a0 6= 0, then the Maclaurin expansion

1 2 n n−1 = c0 + c1D + c2D + . . . anD + an−1D + · · · + a1D + a0 1 where ck = 0 for k < 0, c0 = , ck = q · (ck−n, ck−n+1,...,ck−1), for a0 1 k = 1, 2,... is a dot product of vectors q = (−an, −an−1,..., −a1), a0 (ck−n, ck−n+1,...,ck−1). Proof. The statement follows from the identity n n−1 2 1= anD + an−1D + · · · + a1D + a0 c0 + c1D + c2D + . . . (5.18) after expanding the right-hand side of (5.18) and comparing coeﬃcients of the same powers of D we get 1 c0 = , a0 The coeﬃcient of Dk, k = 1, 2, 3,... we get from the equation

cka0 + ck−1a1 + ck−2a2 + . . . + ck−n+1an−1 + ck−nan = 0, where ck = 0 for k < 0. From this, we immediately have the proof of the Theorem 6. Example 10. Find a particular solution of the equation (D3 − D2 + 2D + 1)y = x3 + 2x2 + 3x (5.19) using the Maclaurin expansion of the inverse diﬀerential operator. MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 29

Solution. Compute the coeﬃcients of the Maclaurin expansion

c0 = 1

c1 = (−1, 1, −2) · (c−2, c−1, c0) = (−1, 1, −2) · (0, 0, 1) = −2

c2 = (−1, 1, −2) · (c−1, c0, c1) = (−1, 1, −2) · (0, 1, −2) = 5

c3 = (−1, 1, −2) · (c0, c1, c2) = (−1, 1, −2) · (1, −2, 5) = −13

c4 = (−1, 1, −2) · (c1, c2,c3) = (−1, 1, −2) · (−2, 5, −13) = 33 . . Then we have 1 (x3 + 2x2 + 3x) D3 − D2 + 2D + 1 = (1 − 2D + 5D2 − 13D3 + 33D4 + · · · )(x3 + 2x2 + 3x) = (1 − 2D + 5D2 − 13D3)x3 + (1 − 2D + 5D2)2x2 + (1 − 2D)3x = (x3 − 6x2 + 30x − 78) + (2x2 − 8x + 20) + (3x − 6) = x3 − 4x2 + 25x − 64 The particular solution of the diﬀerential equation (5.19) y = x3 − 4x2 + 25x − 64 We will also present a shortened calculation using a matrix diﬀerential operator. The solution of (5.19) belongs to the vector space V = span(x3,x2, x, 1) with basis B = {x3,x2, x, 1}. Then 0 00 0 00 00 0 00 0 3 00 0 00 00 0 00 0 D = , D2 = , D3 = B 0 20 0 B 60 00 B 0 00 0 0 01 0 02 00 6 00 0             00 00 00 00 Dn = , for n> 3 B 00 00 00 00    2  3 3 2 yB = (I4 − 2DB + 5DB − 13DB)[x + 2x + 3x]B 1 0 00 1 1 −6 1 00 2 −4 = =  30 −4 1 0 3  25  −78 10 −2 1 0 −64       The particular solution      y = x3 − 4x2 + 25x − 64 30 JOZEF FECENKO

Example 11. Using block matrices determine a particular solution of the differential equation − − (D2 + 6D + 13)y = 2xe 3x sin 2x − 4xe 3x cos 2x (5.20) Solution. Since the roots of the characteristic equation are −3 + 2i, −3 − 2i compared to the right-hand side of the diﬀerential equation, it follows that the particular solution will be in a vector space − − − − V =span x2e 3x sin 2x,x2e 3x cos 3x,xe 3x sin 2x,xe 3x cos 3x, −3x −3x e sin 2x, e cos 3x with basis B of V − − − − B = x2e 3x sin 2x,x2e 3x cos 3x,xe 3x sin 2x,xe 3x cos 3x, −3x −3x e sin 2x, e cos 3x

−3 −2 0 0 0 0 2 −3 0 0 0 0   2 0 −3 −2 0 0 D = B  0 2 2 −3 0 0     0 0 1 0 −3 −2     0 0 0 1 2 −3    We express this matrix as a block matrix  C 0 0 DB = 2I C 0  0 I C   where −3 −2 1 0 0 0 C = , I = , 0 = 2 −3 0 1 0 0 Using mathematical induction it can be proved that Cn 0 0 n n−1 n DB = 2nC C 0 , n = 0, 1, 2,... (5.21)  n− −  n(n − 1)C 2 nCn 1 Cn   It is easy verify that equation (5.21) is true also for n = −1, −2,..., i.e. C−n 0 0 −n −n−1 −n DB = −2nC C 0 n = 1, 2, 3,...  − − − − −  n(n + 1)C n 2 −nC n 1 C n   MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 31

Then I 0 0 C2 0 0 C 0 0 I 0 0 2 2 D +6DB + 13 0 I 0 = 4C C 0 +6 2I C 0 + 13 0 I 0 B         0 0 I 2I 2C C2 0 I C 0 0 I C2 +6C + 13 I  0  0  0 00 = 4C + 12I C2 +6C + 13I 0 = 4C + 12I 0 0     2I 2C +6I C2 +6C + 13I 2I 2C +6I 0     because 5 12 −3 −2 1 0 0 0 C2 + 6C + 13I = + 6 + 13 = = 0 −12 5 2 −3 0 1 0 0 Next 0 −4 2C + 6I = 4 0 0 −8 4C + 12I = 2(2C + 6I)= 8 0 2 0 2I = 0 2 Dividing the matrix into 2 × 2 blocks, we have 0 0 0 4C + 12I 0 0  2I 2C + 6I 0    From Theorem 4, we have

+ 0 0 0 b1 y = 4C + 12I 0 0 b B    2 2I 2C + 6I 0 b3  1 −1    0 2 (2C + 6I) 0 b1 = 0 −(2C + 6I)−2 (2C + 6I)−1 b (5.22)    2 0 0 0 b3  1 −1    2 (2C + 6I) b2 −2 = −(2C + 6I) b2  b∗    0 2 0 ∗ 0 where b = , b = , b = , b = . 1 0 2 −4 3 0 0 32 JOZEF FECENKO

Let us verify that the product of the following block matrices is the identity matrix

1 4C + 12I 0 (2C + 6I)−1 0 2 2I 2C + 6I −2 −1 "−(2C + 6I) (2C + 6I) # 1 (4C + 12I) (2C + 6I)−1 0 2 =  1  2I (2C + 6I)−1 − (2C + 6I(2C + 6I)−2 (2C + 6I)(2C + 6I)−1  2   I 0 I 0  = − − = (2C + 6I) 1 − (2C + 6I) 1 I 0 I

The same is true in reverse order of multiplying the matrices. Let us compute the elements of the last matrix in (5.22)

1 1 −1 1 − 1 0 −4 2 1 0 2 − (2C + 6I) 1b = = 4 = 2 2 2 2 4 0 −4 2  1  −4  1 − 0 −  4   4    

1 2 −2 − 0 −4 2 0 2 −(2C + 6I) 2b = − = − 4 2 4 0 −4  1  −4 − 0  4  1  1  − 0 2 = − 16 = 8  1  −4  1 0 − −  16  4    

0 0b + 0b + 0b = 1 2 3 0

From this we can ﬁnd the particular solution MATRIX DIFFERENTIAL OPERATOR METHOD FOR PARTICULAR SOLUTION ODE 33

1 1 − −  2   2 1 1  −  −   4   4 1 −  1   1  (2C + 6I) 1b     2 2     y = −2 =  8  =  8  B −(2C + 6I) b2     ∗  1   1 b  −  −     4   4            0   0                   0   0      or    

1 − 1 − 1 − 1 − y = − x2e 3x sin 2x − x2e 3x cos 2x + xe 3x sin 2x − xe 3x cos 2x 2 4 8 4 Although the original matrix DB was 6×6, during the calculation of particular solution of the diﬀerential equation (5.20) we calculated the 2 × 2 inverse matrix only.

6. Conclusion The operational method is a fast and universal mathematical tool for ob- taining solutions of differential equations. Determining particular solutions of ordinary nonhomogeneous linear differential equations with constant coefficients using the undetermined coefficients method and the differential operator method are generally known.In particular, distinct from the differential operator method introduced in the literature, we propose and highlight utilizing the definition of the pseudoinverse matrix differential operator to determine a particular solution of differential equations. This method is simple to understand and to apply, compared to some cases of the differential operator method. However, it requires the determination of an inverse or pseudoinverse matrix, which generally has a special type. If the matrix is singular, then we will take a pseudoinverse matrix - Moore Penrose pseudoinverse matrix - instead of the inverse matrix. The paper shows that for its determination, we need to calculate only the inverse submatrix of the considered matrix. Finally, the technique of calculating a particular solution by expressing a matrix differential operator as a block matrix is illustrated. Combination of the operational method, the method of undetermined coefficients and application of the matrix differential method provides a powerful instrument to determine particular solutions of differential equations. 34 JOZEF FECENKO

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University of Economics, Faculty of Economic Informatics, Department of Mathematics and Actuarial Science, Dolnozemska´ St., 832 04 Bratislava, Slo- vakia Email address: [email protected]