a taylor collocation method for the numerical solution of complex differential equations with mixed...

Applied Mathematics and Computation 182 (2006) 498–508

www.elsevier.com/locate/amc

A Taylor collocation method for the numerical solutionof complex differential equations with mixed conditions

in elliptic domains

Mehmet Sezer *, Mustafa Gulsu, Bekir Tanay

Department of Mathematics, Faculty of Science, Mugla University, 48000 Mugla, Turkey

Abstract

An approximate method for solving higher-order linear complex differential equations in the elliptic domains is pro-posed. The approach is based on a Taylor collocation method, which consists of the matrix representation of expressionsin the differential equation and the collocation points defined in the elliptic domain. Illustrative examples are included todemonstrate the validity and applicability of the technique, and performed on the computer using a program written inMaple9.� 2006 Elsevier Inc. All rights reserved.

Keywords: Complex differential equations; Taylor polynomials and series; Taylor collocation methods

1. Introduction

When a mathematical model is formulated for a physical problem it is often represented by complex dif-ferential equations that are not solvable exactly by analytic techniques. Therefore one must resort to approx-imation and numerical methods. For example, the vibrations of a one-mass system with two degrees offreedom are mostly described using differential equation with a complex dependent variable. The differentialequation is usually linear as is shown in papers [1,2]. The solution of the differential equation clarifies the linearphenomena which occur in the system. The various methods for solving differential equations with complexdependent variable are introduced [2].

In recent years, the studies on complex differential equations, i.e., a geometric approach based on meromor-phic function in arbitrary domains [3], a topological description of solution of some complex differential equa-tions with multivalued coefficients [4], the zero distribution [5] and growth estimates [6] of linear complexdifferential equations, the rational and polynomial approximations of analytic functions in the complex plane[7,8], are developed very rapidly and intensively.

0096-3003/$ - see front matter � 2006 Elsevier Inc. All rights reserved.

doi:10.1016/j.amc.2006.04.011

* Corresponding author.E-mail addresses: [email protected] (M. Sezer), [email protected] (M. Gulsu).

mailto:[email protected]

mailto:[email protected]

M. Sezer et al. / Applied Mathematics and Computation 182 (2006) 498–508 499

On the other hand, some Taylor and Chebyshev (matrix and collocations) methods to solve linear differ-ential, integral, integro-differential, difference and integro-difference equations, have been presented in manypaper by Sezer et al. [9–16]. Our purpose in this study is to develop and to apply the mentioned methods aboveto the linear complex differential equation:
Xm
k¼0

P kðzÞf ðkÞðzÞ ¼ gðzÞ; ð1Þ

which is a generalized case of the complex differential equations given in [5,6,17–19], with the mixedconditions:

Xm�1

k¼0

XJ

j¼0

arkf ðkÞðnjÞ� �

¼ kr; r ¼ 0; 1; . . . ;m� 1 ð2Þ

and to find the solution in terms of the Taylor polynomial form, at the point z = 0,

f ðzÞ ¼XN

n¼0

fnzn; f n ¼f ðnÞð0Þ

n!; z 2 D; N P m: ð3Þ

Here Pk(z) and g(z) are analytical functions in the elliptic domain

D ¼ z ¼ xþ iy; z 2 C; x; y 2 R; x ¼ a cos h; y ¼ b sin h;f0 < h 6 2p;�a 6 x 6 a;�b 6 y 6 b; a; b 2 Rþg;

ark and kr are appropriate complex or real constants; ni 2 D; the coefficients fn, n = 0,1, . . .,N are the Taylorcoefficients to be determined.

2. Determination of collocation points in elliptic domains

We first define the collocation points to be used in the solution method by

zpq ¼ xpq þ iypq ¼ ðxpq; ypqÞ 2 D;

xpq ¼aN

p coshN

q; ypq ¼bN

p sinhN

qð4Þ

so that

p; q ¼ 0; 1; . . . ;N ; 0 < h 6 2p; z0q ¼ z00 ¼ ð0; 0Þ:
Note that for h = 2p, p = 0, q = 0,1, . . .,N, z0q = z00 = 0 + 0i = (0,0); and for
h ¼ 2p; p 6¼ 0; q ¼ 0 and q ¼ N ;

zp0 = zpN; that is z10 = z1N, z20 = z2N, . . .,zN0 = zNN.For example, the collocation points for a = 6, b = 3, N = 3, h = 2p are given in Fig. 1.Here for q = 0, q = 3 and p = 0,1,2,3,

z00 ¼ z01 ¼ z02 ¼ z03 ¼ ð0; 0Þ; z10 ¼ x10 þ iy10 ¼ ð2; 0Þ ¼ z13;

z20 ¼ x20 þ iy20 ¼ ð4; 0Þ ¼ z23; z30 ¼ x30 þ iy30 ¼ ð6; 0Þ ¼ z33;

for q = 1 and p = 1, 2, 3,

z11 ¼ x11 þ iy11 ¼ �1;

ffiffiffi3p

2

!; z21 ¼ x21 þ iy21 ¼ ð�2;

ffiffiffi3pÞ; z31 ¼ x31 þ iy31 ¼ �3;

3ffiffiffi3p

2

!;

for q = 2 and p = 1, 2, 3,

z12 ¼ x12 þ iy12 ¼ �1;�ffiffiffi3p

2

!; z22 ¼ x22 þ iy22 ¼ ð�2;�

ffiffiffi3pÞ; z32 ¼ x32 þ iy32 ¼ �3;

�3ffiffiffi3p

2

!:

y

z00 z10 z20

z11

z21

z31

z12

z22

z32

z30

x

Fig. 1. Collocation points for a = 6, b = 3, N = 3, h = 2p.

500 M. Sezer et al. / Applied Mathematics and Computation 182 (2006) 498–508

Another example, the collocation points for a = 6, b = 3, N = 4, h = 2p are

z0q ¼ x0q þ iy0q ¼ ð0; 0Þ; q ¼ 0; 1; 2; 3; 4;

z10 ¼ x10 þ iy10 ¼6

4; 0

� �¼ z14; z20 ¼ x20 þ iy20 ¼

12

4; 0

� �¼ z24;

z30 ¼ x30 þ iy30 ¼18

4; 0

� �¼ z34; z40 ¼ x40 þ iy40 ¼

24

4; 0

� �¼ z44;

z11 ¼ x11 þ iy11 ¼ 0;3

4

� �; z12 ¼

�6

4; 0

� �; z13 ¼ 0;

�3

4

� �;

z21 ¼ x21 þ iy21 ¼ 0;6

4

� �; z22 ¼

�12

4; 0

� �; z13 ¼ 0;

�6

4

� �;

z31 ¼ x31 þ iy31 ¼ 0;9

4

� �; z32 ¼ ð

�18

4; 0Þ; z13 ¼ 0;

�9

4

� �;

z41 ¼ x41 þ iy41 ¼ 0;12

4

� �; z42 ¼

�24

4; 0

� �; z13 ¼ 0;

�12

4

� �:

In the case h = p, N = 3 a = 6 and b = 3, the collocation points become,

zpq ¼ xpq þ iypq; p; q ¼ 0; 1; 2; 3;

xpq ¼ 2p cosp3

q; ypq ¼ p sinp3

q;

so that

z0q ¼ ð0; 0Þ ¼ z00:

In the case h = p, N = 4, a = 6 and b = 3 the collocation points become,

zpq ¼ xpq þ iypq; p; q ¼ 0; 1; 2; 3; 4;

xpq ¼3

2p cos

p4

q; ypq ¼3

4p sin

p4

q;

so that

z0q ¼ ð0; 0Þ ¼ z00:


3. Fundamental matrix relations

We first consider the solution f(z) and its derivative f(k)(z) in the forms

f ðzÞ ¼ f ð0ÞðzÞ ¼XN

n¼0


n!

and

f ðkÞðzÞ ¼XN

n¼0

f ðkÞn zn: ð5Þ

It is well known from [16] that the relation between the coefficients f ðkþ1Þn and f ðkÞnþ1 is

f ðkþ1Þn ¼ ðnþ 1Þf ðkÞnþ1; n; k ¼ 0; 1; 2; . . . ð6Þ

Then we convert the expressions in (5) to matrix forms, respectively,

½f ðzÞ� ¼ ZðzÞF ð7Þ
and
f ðkÞðzÞ� �

¼ ZðzÞFðkÞ ð8Þ

so that

ZðzÞ ¼ 1 z z2 � � � zN� �

;

F ¼ ½ f0 f1 � � � fN �T;

FðkÞ ¼ ½ f ðkÞ0 f ðkÞ1 � � � f ðkÞN �T:

Note that

F ¼ Fð0Þ ¼ ½ f ð0Þ0 f ð0Þ1 � � � f ð0ÞN �T:

In addition, from the recurrence relation (6), it is obtained the matrix relation [16],

Fðkþ1Þ ¼MFðkÞ; k ¼ 0; 1; . . .

or

FðkÞ ¼MkF; ð9Þ
where
M ¼

0 1 0 � � � 0 0

0 0 2 � � � 0 0

� � � � � � � � � �� 0 0 0 � � � N

0 0 0 � � � 0 0

2666666666664

3777777777775; M0 ¼

1 0 0 � � � 0 0

0 1 0 � � � 0 0

� � � � � � � � � �� 0 0 0 � � � 0

0 0 0 � � � 0 1

2666666666664

3777777777775:

Substituting (9) into (8) we have the relation

f ðkÞðzÞ� �

¼ ZðzÞMkF: ð10Þ

For the collocation points z = zpq defined by (4), the matrix relation (10) becomes:

f ðkÞðzpqÞ� �

¼ ZðzpqÞMkF; ð11Þ


where

ZðzpqÞ ¼ ½ 1 zpq z2pq � � � zN

pq �; p; q ¼ 0; 1; . . . ;N :

For p = 0,1, . . .,N we can write the relation (11) in the form

f ðkÞðz0qÞ� �

¼ Zðz0qÞMkF;

f ðkÞðz1qÞ� �

¼ Zðz1qÞMkF;

� � �f ðkÞðzNqÞ� �

¼ ZðzNqÞMkF;

or briefly

FðkÞq ¼

f ðkÞðz0qÞf ðkÞðz1qÞ

..

.

f ðkÞðzNqÞ

2666664

3777775 ¼ ZqMkF; q ¼ 0; 1; . . . ;N ; ð12Þ

where

Zq ¼

Zðz0qÞZðz1qÞ

..

.

ZðzNqÞ

266664

377775 ¼

1 z0q z20q � � � zN

0q

1 z1q z21q � � � zN

1q

� � � �� 1 zNq z2

Nq � � � zNNq

26666666664

37777777775:

On the other hand, substituting the collocation points z = zpq into Eq. (1) we have,

Xm

k¼0

P kðzpqÞf ðkÞðzpqÞ ¼ gðzpqÞ; p; q ¼ 0; 1; . . . ;N : ð13Þ

By using the expressions (11)–(13), we obtain the fundamental matrix equation,

Xm

k¼0

XN

q¼0

PkqZqMkF ¼XN

q¼0

Gq; ð14Þ

so that

Gq ¼ ½ gðz0qÞ gðz1qÞ � � � gðzNqÞ �T

and

Pkq ¼

P kðz0qÞ 0 � � � 0

0 P kðz1qÞ � � � 0

� � �� 0 0 � � � P kðzNqÞ

2666666664

3777777775:


We can obtain the corresponding matrix form to the conditions (2) as follows. By means of the relation (10)we have the matrix equation,

Xm

k¼0

XJ

j¼0

arkZðnjÞMk

( )F ¼ ½kr�; r ¼ 0; 1; . . . ;m� 1; ð15Þ

where

ZðnjÞ ¼ ½ 1 nj n2j � � � nN

j �:

Briefly, the system of the matrix Eq. (15) can be written in the matrix form,

UrF ¼ ½kr�; ð16Þ
where
Ur ¼Xm�1

k¼0

XJ

j¼0

arkZðnjÞMk � ½ ur0 ur1 � � � urN �; r ¼ 0; 1; . . . ;m� 1:

4. Method of solution

We now consider the fundamental matrix Eq. (14) corresponding to Eq. (1). We can write Eq. (14) in theform,

WF ¼ G; ð17Þ
where
W ¼ ½wst� ¼Xm

k¼0

XN

q¼0

PkqZqMk; s; t ¼ 0; 1; . . . ;N

and

G ¼XN

q¼0

Gq � ½ g0 g1 � � � gN �T

Gq is defined in (14). The augmented matrix of Eq. (17) becomes

½W; G� ¼ ½wst; gs�; s; t ¼ 0; 1; . . . ;N : ð18Þ
The augmented matrix of the Eq. (16) corresponding to condition (2) can be written in the form,
½Ur; kr� ¼ ½ ur0 ur1 � � � urN; kr � ð19Þ

where Ur is defined in (16).Consequently, to find the unknown Taylor coefficients fn, n = 0,1, . . .,N, related with the approximate solu-

tion of the problem consisting of Eq. (1) and conditions (2), by replacing the m row matrices (19) by the last m

rows of the augmented matrix (18) [9], we have new augmented matrix,

½W�; G�� ¼

w00 w01 � � � w0N ; g0

w10 w11 � � � w1N ; g1

� � � � � � � � � ; � � �wN�m;0 wN�m;1 � � � wN�m;N ; gN�m

u00 u01 � � � u0N ; k0

u10 u11 � � � u1N ; k1

� � � � � � � � � ; � � �um�1;0 um�1;1 � � � um�1;N ; km�1

266666666666664

377777777777775

ð20Þ


or the corresponding matrix equation

W�F ¼ G�:

If detW* 5 0, we can write Eq. (20) as,

F ¼ W�ð Þ�1G�

and the matrix F is uniquely determined. Thus the mth-order linear complex differential equation with variablecoefficients (1) under the conditions (2) has a unique solution in the form (3).

Also we can easily check the accuracy of the obtained solutions as follows [13,15].Since the Taylor polynomial (3) is an approximate solution of Eq. (1), when the solutions f(z) and its deriv-

atives are substituted in Eq. (1), the resulting equation must be satisfied approximately; that is, for z = zj 2 D,j = 0,1,2, . . .,

EðzjÞ ¼Xm

k¼0

P kðzjÞf ðkÞðzjÞ � gðzjÞ��

�� ffi 0

or

EðzjÞ 6 10�kj ðkj is any positive integerÞ:
If max ð10�kiÞ ¼ 10�k (k is any positive integer) is prescribed, then the truncation limit N is increased until thevalues E(zj) at each of the points zj becomes smaller than the prescribed 10�k.
5. Illustrative examples

In this section, several numerical examples are given to illustrate the properties of the method and all ofthem were performed on the computer using a program written in Maple9. The absolute errors in tablesare the values of jf(z) � fN(z)j at selected points.

Example 1 [16, p. 299]. Let us first consider the second order complex differential equation (see Fig. 2):

f 00ðzÞ þ zf ðzÞ ¼ ez þ zez;

with f(0) = 1, f 0(0) = 1 and approximate the solution f(z) by the truncated Taylor series in the form:

f ðzÞ ¼X4

n¼0


n!;

so that a = 1, b = 1/2, P0(z) = z, P1(z) = 0, P2(z) = 1 and g(z) = ez + zez. For N = 4, and h = p we have thecollocation points

0.5

1

1.5

2

2.5

3

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1

N=6

exact

Fig. 2. Numerical and exact solution of the Example 1 for N = 6.

TableError

z

0.0 + 00.1 + 00.2 + 00.3 + 00.4 + 00.5 + 00.6 + 00.7 + 00.8 + 00.9 + 01.0 + 1

TableError

z

0.0 + 00.1 + 00.2 + 00.3 + 00.4 + 00.5 + 00.6 + 00.7 + 00.8 + 00.9 + 01.0 + 1


0 0 0 0 014

ffiffi2p

8þffiffi2p

16i 1

8i �

ffiffi2p

8þffiffi2p

16i �1

4

12

ffiffi2p

4þffiffi2p

8i 1

4i �

ffiffi2p

4þffiffi2p

8i � 1

2

34

3ffiffi2p

8þ 3

ffiffi2p

16i 3

8i � 3

ffiffi2p

8þ 3

ffiffi2p

16i �3

4

1ffiffi2p

2þffiffi2p

4i 1

2i �

ffiffi2p

2þffiffi2p

4i �1

8>>>>>>><>>>>>>>:

9>>>>>>>=>>>>>>>;

and the fundamental matrix equation

X2

k¼0

X4

q¼0

PkqZqMkF ¼X4

q¼0

Gq:

This equation has the solution

F ¼ ½ 1 1 0:5 0:1642306718þ 0:001125922559i 0:04292618462þ 0:003526627844i �T:

Therefore, we find the solution

f ðzÞ ¼ 1þ zþ 0:5z2 þ ð0:1642306718þ 0:001125922559iÞz3 þ ð0:04292618462þ 0:003526627844iÞz4:

The values of this solution are compared with the exact solution f(z) = ez for N = 5, 6 in Tables 1–3.

1analysis of Example 1 for the Re(z) value

Exact solution (Real) Present met (z0 = 0)

N = 5 (Real) Ne = 5 N = 6 (Real) Ne = 6

.0i 1.00000000 1.00000000 0.0000000 1.00000000 0.0000000

.1i 1.09964966 1.09965061 0.9500E�6 1.09964966 0.1600E�7

.2i 1.19705602 1.19706225 0.6238E�5 1.19705602 0.5450E�6

.3i 1.28956937 1.28957462 0.5247E�5 1.28956937 0.4566E�5

.4i 1.37406153 1.37400924 0.5229E�4 1.37406153 0.2039E�4

.5i 1.44688903 1.44660517 0.2838E�3 1.44688903 0.6384E�4

.6i 1.50385954 1.50295453 0.9050E�3 1.50385954 0.1583E�3

.7i 1.54020302 1.53793823 0.2264E�2 1.54020302 0.3317E�3

.8i 1.55054929 1.54566154 0.4887E�2 1.55054929 0.6078E�3

.9i 1.52891381 1.51938975 0.9524E�2 1.52891381 0.9909E�3

.0i 1.46869394 1.45148379 0.1721E�1 1.46869394 0.1441E�2

2analysis of Example 1 for the Im(z) value

Exact solution (Im) Present met (z0 = 0)

N = 5 (Im) Ne = 5 N = 6 (Im) Ne = 6

.0i 0.00000000 0.00000000 0.0000000 0.00000000 0.0000000

.1i 0.11033298 0.11033432 0.1336E�5 0.11033292 0.6510E�7

.2i 0.24265526 0.24267093 0.1566E�4 0.24265467 0.5889E�6

.3i 0.39891055 0.39899230 0.8175E�4 0.39890858 0.1968E�5

.4i 0.58094390 0.58123963 0.2957E�3 0.58094018 0.3712E�5

.5i 0.79043908 0.79129120 0.8521E�3 0.79043653 0.2544E�5

.6i 1.02884566 1.03094076 0.2095E�2 1.02885694 0.1127E�4

.7i 1.29729511 1.30187588 0.4580E�2 1.29735515 0.6003E�4

.8i 1.59650534 1.60565633 0.9150E�2 1.59669305 0.1877E�3

.9i 1.92667330 1.94369244 0.1701E�1 1.92714583 0.4725E�3

.0i 2.28735528 2.31722346 0.2986E�1 2.28839852 0.1043E�2

Table 3The absolute error of the considered method at various points for Example 1

z Exact solution ði ¼ffiffiffiffiffiffiffi�1p

Þ Ne = 4 Ne = 5 Ne = 6

0.0 + 0.0i 1.000000000 + 0.0000000000i 0.000000000 0.000000000 0.0000000000.1 + 0.1i 1.099649667 + 0.1103329887i 0.854850E�5 0.163957E�5 0.670373E�70.2 + 0.2i 1.197056021 + 0.2426552686i 0.721149E�4 0.168632E�4 0.802389E�60.3 + 0.3i 1.289569374 + 0.3989105538i 0.243685E�3 0.819207E�4 0.497217E�50.4 + 0.4i 1.374061539 + 0.5809439008i 0.570193E�3 0.300320E�3 0.207311E�40.5 + 0.5i 1.446889037 + 0.7904390832i 0.117417E�2 0.898160E�3 0.638936E�40.6 + 0.6i 1.503859541 + 1.0288456667i 0.246267E�2 0.228221E�2 0.158761E�30.7 + 0.7i 1.540203025 + 1.2972951132i 0.527614E�2 0.511006E�2 0.337166E�30.8 + 0.8i 1.550549297 + 1.5965053441i 0.108800E�2 0.103745E�1 0.636201E�30.9 + 0.9i 1.528913812 + 1.9266733004i 0.210188E�1 0.195027E�1 0.109785E�21.0 + 1.0i 1.468693940 + 2.2873552867i 0.380390E�1 0.344716E�1 0.177916E�2


Example 2. Consider the linear second order complex differential equation:

TableThe ab

z

0.0 + 00.1 + 00.2 + 00.3 + 00.4 + 00.5 + 00.6 + 00.7 + 00.8 + 00.9 + 01.0 + 1

ð1� z2Þf 00ðzÞ � 2zf 0ðzÞ þ 6f ðzÞ ¼ z2 sin z� 2z cos zþ 5 sin z� 3;

with f(0) = �1, f 0(0) = 1 and approximate the solution f(z) by the Taylor polynomial,

f ðzÞ ¼X6

n¼0


n!;

so that a = 1, b = 1/2, P0(z) = 6, P1(z) = �2z, P2(z) = 1 � z2, h = p and

gðzÞ ¼ z2 sin z� 2z cos zþ 5 sin z� 3:

It can be seen that the exact solution of this problem is f ðzÞ ¼ 32z2 þ sin z� 1.

When the presented method is applying to the problem, the fundamental matrix equation becomes:

X2

k¼0

X6

q¼0

PkqZqMkF ¼X6

q¼0

Gq:

Hence, The comparison of the absolute errors with different N is given in Table 4.

Example 3. Our last example is the linear complex differential equation:

f 000ðzÞ þ 2f 00ðzÞ þ f 0ðzÞ þ f ðzÞ ¼ 2z2 þ 6zþ 11;

4solute error of the considered method at various points for Example 2

Exact solution ði ¼ffiffiffiffiffiffiffi�1p

Þ Ne = 5 Ne = 6 Ne = 7

.0i �1.0000000000 + 0.0000000000i 0.000000000 0.000000000 0.00000000

.1i �0.8996670002 + 0.1296663335i 0.166725E�6 0.135844E�7 0.145845E�8

.2i �0.7973440203 + 0.3173226870i 0.160118E�5 0.107731E�5 0.237736E�7

.3i �0.6910813463 + 0.5609193480i 0.689432E�5 0.666524E�5 0.186524E�6

.4i �0.5790105890 + 0.8583279455i 0.228819E�4 0.182569E�4 0.492563E�5

.5i �0.4593873143 + 1.2073041532i 0.655500E�4 0.604268E�4 0.624268E�4

.6i �0.3306359870 + 1.6054528761i 0.166363E�3 0.337509E�3 0.527501E�4

.7i �0.1913979261 + 2.0501968182i 0.382185E�3 0.242106E�3 0.832109E�4

.8i �0.0405828954 + 2.5387493965i 0.811074E�3 0.111662E�3 0.661666E�3

.9i 0.12257513052 + 3.068093029i 0.161540E�2 0.112954E�2 0.712957E�3

.0i 0.29845758112 + 3.634963915i 0.305327E�2 0.282865E�2 0.492861E�3


with the condition f(0) = 1, f 0(0) = 2, f00(0) = 4. For N = 4, a = 1, b = 1/2 and h = p the matrix form of theproblem is defined by:

X3

k¼0

X4

q¼0

PkqZqMkF ¼X4

q¼0

Gq:

After the augmented matrices of the systems and conditions are computed, we obtain the solution:

F ¼ ½ 1 2 2 0 0 �T:

Therefore, we find the exact solution:

f ðzÞ ¼ 2z2 þ 2zþ 1:

6. Conclusions

High order linear complex differential equations are usually difficult to solve analytically. Then it is requiredto obtain the approximate solutions. For this reason, the present method has been proposed for approximatesolution and also analytical solution.

The method presented in this study is a method for computing the coefficients in the Taylor expansion ofthe solution of a linear complex differential equations, and is valid when the functions Pk(z) and g(z) aredefined in the elliptic domain D = {z 2 C, z = x + iy, x = acosh, y = b sinh, 0 6 h 6 2p, �a 6 x 6 a,� b 6 y 6 b, a, b 2 R}. The Taylor method is an effective method for the cases that the known functions havethe Taylor series expansion at z = z0. In this case, the Taylor polynomial solution f(z) and the values f(zj),zj 2 D can be easily evaluated at low-computation effort. In addition, an interesting feature of this methodis to find the analytical solutions if the equation has an exact solution that is a polynomial of degree N or lessthan N.

The method can also be extended to the system of linear complex differential equations with variable coef-ficients, but some modifications are required.

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