ABSTRACT 

 Ordinary differential equations, both first and second order, are essential in the modeling 

of many physical systems. A system of simultaneous differential equations results from more 

complicated modeling involving more than one dependent variables with respect to a single 

independent variable. There are several methods in solving a system of simultaneous linear 

differential equations including variable substitution, Laplace transform and using the D-operator.  

 Proposed in this paper is a simplified method of solving a set of two non-homogeneous 

linear first-order simultaneous ordinary differential equations with constant coefficients that falls 

into a certain form. An algebraic formula is developed to compute the solution to the said 

differential equations provided a certain necessary condition is satisfied. 

 Four different forms of the functions of the independent variable on the right side of the 

equations, namely constants, linear functions, natural exponential functions and sinusoidal 

functions, are considered. For each case, an algebraic formula to calculate the dependent variable 

as well as its derivative are developed. Moreover, these simple algebraic formulae can be easily 

programmed into spreadsheet where one just has to enter the values of the constants and 

coefficients from the original equations and instantly obtain the correct answers. 

  

Key words: Simultaneous differential equations; Non-homogeneous differential equations; 

Cramer’s rule. 

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  Algebraic Solution to Simultaneous Linear First-Order Non-

Homogeneous Differential Equations with Constant Coefficients

Seng Chu Chow

INTI International College Subang, Malaysia
 



 
 

 

 

1. Introduction 

  Ordinary Differential Equations (ODE) play a crucial role in understanding and modeling 

various physical phenomena across many fields such as engineering, sciences and economics. 

Among the numerous applications of first-order differential equations are solutions mixing, 

population change, heating or cooling, free-falling body, fluid dynamics, resistive-capacitive (RC) 

and resistive-inductive (RL) circuits. Second-order differential equations find its applications in 

spring-mass vibration, RLC circuits, wave propagations, etc. By solving a differential equation 

using its known initial conditions, which is known as Initial Value Problem (IVP), we can 

anticipate the behavior of the physical system over time. [7] 

  Simultaneous differential equations is a system of at least two differential equations with 

two or more dependent variables but share a single common independent variable. These 

equations govern the inter-relationship between the rate of change of the dependent variables 

with respect to the independent variable. [8] 

 The general form of an nth-order linear Ordinary Differential Equation is: 

 𝑎𝑛(𝑥)
𝑑𝑛𝑦

𝑑𝑥𝑛 + 𝑎𝑛−1(𝑥)
𝑑𝑛−1𝑦

𝑑𝑥𝑛−1 + ⋯ + 𝑎1(𝑥)
𝑑𝑦

𝑑𝑥
+ 𝑎0(𝑥)𝑦 = 𝐹(𝑥)            (1) 

 Since this paper only consider 1st-order ODE, equation (1) reduces to 

 𝑎1(𝑥)
𝑑𝑦

𝑑𝑥
+ 𝑎0(𝑥)𝑦 = 𝐹(𝑥)                 (2) 

 where  𝑎1(𝑥), 𝑎2(𝑥) and 𝐹(𝑥) are functions of x only, and there is no restriction on the nature 

of these x-dependencies. [5, 6] 

 

     

   

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2   Example of a System of Simultaneous 1st-order ODE

2.1 Double Mixing Problem



 
 

  Figure 1 below shows a brine solution with concentration 𝐵1(𝑘𝑔/𝑙) flows at a constant 

rate of 𝐹1(𝑙/𝑠) into Tank 1 that initially contains 𝑉1(𝑙) of the same brine. The solution in the 

tank is kept well stirred and flows out of Tank 1 at a rate of 𝐹2 with concentration 𝐵2 directly 

into Tank 2 with volume 𝑉2. The solution in Tank 2 is also well mixed and flows out at a rate of 

𝐹3 with concentration 𝐵3 . The mass of salt in Tank 1 and Tank 2 are 𝑚1(𝑡)  and 𝑚2(𝑡) 

respectively.  

 

 

  The IVP for this problem is given by: 

   
𝑑𝑚1

𝑑𝑡
= 𝐵1𝐹1 −

𝑚1

𝑉1+(𝐹1−𝐹2)∙𝑡
𝐹2   ,   𝑚1(0) = 𝑎            (3a) 

   
𝑑𝑚2

𝑑𝑡
=

𝑚1

𝑉1+(𝐹1−𝐹2)∙𝑡
𝐹2 −

𝑚2

𝑉2+(𝐹2−𝐹3)∙𝑡
𝐹3   ,   𝑚2(0) = 𝑏                  (3b) 

 where  a and b are the initial mass of salt in Tank 1 and Tank 2, respectively. 

 

 2.2 Simplifying IVP  

  In most applications, 𝐹1, 𝐹2 and 𝐹3 are constant and equal, otherwise the volume of liquid 

in the two tanks would be changing with time. If the two volumes are assumed to be equal, 

that is 𝑉1 = 𝑉2 = 𝑉, then the IVP in equation (3) is simplified to 

   𝑚1
′ = 𝐴 − 𝐾𝑚1              (4a) 

   𝑚2
′ = 𝐾𝑚1 − 𝐾𝑚2              (4b) 

 where 

Figure 1: Double Mixing Problem 

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   𝐴 = 𝐵1𝐹,   𝐾 =
𝐹

𝑉
   and   𝐹1 = 𝐹2 = 𝐹3 = 𝐹.           (4c) 

  Putting equation (4) in matrix form, we have 

   [ 
𝑚1

′

𝑚2
′  ] = [ 

−𝐾 0
𝐾 −𝐾

 ] [ 
𝑚1

𝑚2
 ] + [ 

𝐴
0

 ]              (5) 

  If the input concentration can be continuously adjusted to be in direct proportional to the 

varying mass of salt in Tank 1, that is 𝐵1 ∝ 𝑚1, then 

   𝐴 = 𝑝𝑚1𝐹 = 𝐶 ∙ 𝑚1                 (6) 

 where p is the constant of proportionality between 𝐵1  and 𝑚1 , and the constant c is the 

product of p and F. With this assumption, equation (5) can be simplified to 

   [ 
𝑚1

′

𝑚2
′  ] = [ 

−𝐾 + 𝐶 0
𝐾 −𝐾

 ] [ 
𝑚1

𝑚2
 ]              (7) 

 

 2.3 De-coupling of the System  

  The eigenvalues of equation (7) can be found by solving the characteristic polynomial  

   |
𝜆 + 𝐾 − 𝐶 0

−𝐾 𝜆 + 𝐾
| = 0                (8) 

 which yields 

   𝜆1 = −𝐾   and   𝜆2 = −(𝐾 − 𝐶)              (9) 

 with the corresponding eigenvectors   

   𝛼1 = 𝑟 [ 
0
1

 ]   and   𝛼2 = 𝑠 [
𝐶

 𝐾

1
 ]  ,   𝑟, 𝑠 ∈ 𝑅.            (10) 

 The transition matrix P and its inverse is given by 

   𝑃 = [ 
0

𝐶

𝐾

1 1
 ]   ,   𝑃−1 =

1

−
𝐶

𝐾
 
 [

 1 −
𝐶

𝐾
 

−1   0
]             (11) 

 As expected,  

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1

−
𝐶

𝐾
 
 [

 1 −
𝐶

𝐾
 

−1   0
] [ 

−𝐾 + 𝐶 0
𝐾 −𝐾

 ] [ 
0

𝐶

𝐾

1 1
 ] = [

𝜆1 0
0 𝜆2

]          (12) 

 Let’s define  

   𝑚 = 𝑃𝑥   and 𝑚′ = 𝑃𝑥′               (13) 

  where 𝑥 = [
𝑥1(𝑡)
𝑥2(𝑡)

] is a dummy function. 

 Substitute (13) into (7) yields 

   𝑃𝑥′ = [ 
−𝐾 + 𝐶 0

𝐾 −𝐾
 ] 𝑃𝑥              (14) 

 Multiply (14) by 𝑃−1 gives 

   𝑃−1𝑃𝑥′ = 𝑃−1 [ 
−𝐾 + 𝐶 0

𝐾 −𝐾
 ] 𝑃𝑥             (15) 

 Incorporate equations (9), (11) and (12) into (15), we have  

   [
𝑥1

′ (𝑡)

𝑥2
′ (𝑡)

] = [
𝜆1 0
0 𝜆2

] [
𝑥1(𝑡)
𝑥2(𝑡)

]            (16a) 

 which gives 

   𝑥1
′ = 𝜆1𝑥1             (16b) 

   𝑥2
′ = 𝜆2𝑥2             (16c) 

 The system is now decoupled [9, 10]. 

 

 2.4 Solution to example 

  The solutions to equation (16) are easily obtained using separation of variables method 

as 

   𝑥1(𝑡) = 𝐷1𝑒−𝐾𝑡            (17a) 

   𝑥2(𝑡) = 𝐷2𝑒−(𝐾−𝐶)𝑡            (17b) 

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 Since 𝑚 = 𝑃𝑥, we have 

   [ 
𝑚1

𝑚2
 ] = [ 

0
𝐶

𝐾

1 1
 ] [ 

𝐷1𝑒−𝐾𝑡

𝐷2𝑒−(𝐾−𝐶)𝑡 ] .             (18) 

 With the initial conditions 𝑚1(0) = 𝑎 and 𝑚2(0) = 𝑏, it is easy to verify that the solutions 

to the IVP in (3) are  

   𝑚1(𝑡) = 𝑎𝑒−(𝐾−𝐶)𝑡            (19a) 

   𝑚2(𝑡) = (𝑏 −
𝐾𝑎

𝐶
) 𝑒−𝐾𝑡 +

𝐾𝑎

𝐶
𝑒−(𝐾−𝐶)𝑡          (19b) 

 

3   Existing Solution Methods 

 There are several methods to solve simultaneous differential equations, each has it own 

restrictions on the form of the equations. [11, 12, 13, 14, 15, 16] These methods are: 

 (i) Substitution 

 (ii) Elimination 

 (iii) D-Operator 

 (iv) Laplace transform 

 (v) Numerical recursive calculations 

 (vi) Software package 

 

  

  

 

   

  

 

    

  

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4. Proposed Algebraic Solutions

  A special  case  of simultaneous  differential  equations is  focused  here,  where only  one 

dependent variable y, and one independent variable x, are involved. However, the solution must 

satisfy two non-homogeneous differential equations simultaneously.

The main result of this paper is stated as a theorem below.

Theorem 1

  The solutions to the set of non-homogeneous linear first-order simultaneous ordinary 

differential equations with constant coefficients in the form of



 
 

 𝑎1𝑦′ + 𝑎0𝑦 = 𝑓(𝑥)             (20a) 

 𝑏1𝑦′ + 𝑏0𝑦 = 𝑔(𝑥)              (20b) 

are given by 

 𝑦(𝑥) =  
𝑔(𝑥)𝑎1−𝑓(𝑥)𝑏1

𝑎1𝑏0−𝑎0𝑏1
            (21a) 

 𝑦′(𝑥) =
𝑓(𝑥)𝑏0−𝑔(𝑥)𝑎0 

𝑎1𝑏0−𝑎0𝑏1
            (21b) 

with the necessary condition 

 𝑎1𝑔′(𝑥) − 𝑏1𝑓′(𝑥) = 𝑏0𝑓(𝑥) − 𝑎0𝑔(𝑥).           (21c) 

Proof 

Write the given equations in matric form as  

 [
𝑎1 𝑎0

𝑏1 𝑏0
] [

𝑦′

𝑦
] = [

𝑓(𝑥)
𝑔(𝑥)

]              (22) 

Define the following three determinants 

 |𝐴| = |
𝑎1 𝑎0

𝑏1 𝑏0
| = 𝑎1𝑏0 − 𝑎0𝑏1            (23a) 

 |𝐴1| = |
𝑓(𝑥) 𝑎0

𝑔(𝑥) 𝑏0
| = 𝑏0𝑓(𝑥) − 𝑎0𝑔(𝑥)           (23b) 

 |𝐴2| = |
𝑎1 𝑓(𝑥)
𝑏1 𝑔(𝑥)

| = 𝑎1𝑔(𝑥) − 𝑏1𝑓(𝑥)         (23c) 

Using Cramer’s Rule, the solutions are 

 𝑦′(𝑥) =
𝑏0𝑓(𝑥)−𝑎0𝑔(𝑥)

𝑎1𝑏0−𝑎0𝑏1
   and   𝑦(𝑥) =

𝑎1𝑔(𝑥)−𝑏1𝑓(𝑥)

𝑎1𝑏0−𝑎0𝑏1
          (24) 

However, since 𝑦(𝑥) and 𝑦′(𝑥) are not constants and 𝑦′(𝑥) is obtained by differentiating 𝑦(𝑥) 

with respect to x, the following necessary condition must be satisfied:  

 𝑎1𝑔′(𝑥) − 𝑏1𝑓′(𝑥) = 𝑏0𝑓(𝑥) − 𝑎0𝑔(𝑥)   ∎             (25) 

 

Note that the solution 𝑦(𝑥) can be obtained using simple algebraic computation. 

Now let’s look at the four common functions for 𝑓(𝑥) and 𝑔(𝑥). 

 

4.1 Case 1: 𝒇(𝒙) 𝐚𝐧𝐝 𝒈(𝒙) are constants. 

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𝑓(𝑥) = 𝑓 and 𝑔(𝑥) = 𝑔, where f and g are scalar constants. 

The necessary condition becomes 

 0 = 𝑏0𝑓 − 𝑎0𝑔                  (26) 

and the solutions are 

 𝑦′(𝑥) =
𝑏0𝑓−𝑎0𝑔

𝛼
= 0               (27a) 

   𝑦(𝑥) = 
𝑎1𝑔−𝑏1𝑓

𝛼
             (27b) 

where the constant 𝛼 is defined as 

 𝛼 = 𝑎1𝑏0 − 𝑎0𝑏1                (28) 

Numerical Example 1 

 2𝑦′ + 3𝑦 = 6 

 5𝑦′ − 𝑦 = −2 

𝑎1 = 2, 𝑎0 = 3, 𝑏1 = 5, 𝑏0 = −1, 𝑓 = 6 and 𝑔 = −2. 

𝑏0𝑓 − 𝑎0𝑔 = (−1)(6) − (3)(−2) = −6 + 6 = 0 

𝛼 = (2)(−1) − (3)(5) = −17 

So,  

 𝑦′(𝑥) =
(−1)(6)−(3)(−2)

−17
= 0   

 𝑦(𝑥) =
(2)(−2)−(5)(6)

−17
=

−34

−17
= 2 

 

    

   

 

      

                (29) 

 

  

    

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Check:

Eq 1: 2𝑦′ + 3𝑦 = 2(0) + 3(2) = 6 √

Eq 2: 5𝑦′ − 𝑦 = 5(0) − (2) = −2 √

4.2 Case 2: 𝒇(𝒙) 𝐚𝐧𝐝 𝒈(𝒙) are linear functions of x.

  𝑓(𝑥) = 𝑚1𝑥 + 𝑚0 and 𝑔(𝑥) = 𝑛1𝑥 + 𝑛0.

The necessary condition becomes

𝑎1𝑛1 − 𝑏1𝑚1 = 𝑏0(𝑚1𝑥 + 𝑚0) − 𝑎0(𝑛1𝑥 + 𝑛0)

  = 𝑏0𝑚1𝑥 + 𝑏0𝑚0 − 𝑎0𝑛1𝑥 − 𝑎0𝑛0



 
 

   = (𝑏0𝑚1 − 𝑎0𝑛1)𝑥 + (𝑏0𝑚0 − 𝑎0𝑛0)           (30) 

Equating coefficients of like powers yield the following two necessary conditions. 

 𝑏0𝑚1 − 𝑎0𝑛1 = 0             (31a) 

 𝑎1𝑛1 − 𝑏1𝑚1 = 𝑏0𝑚0 − 𝑎0𝑛0           (31b) 

The solutions are 

 𝑦(𝑥) =
(𝑛1𝑥+𝑛0)𝑎1−(𝑚1𝑥+𝑚0)𝑏1

𝛼
 

  =
(𝑎1𝑛1−𝑏1𝑚1)𝑥+(𝑎1𝑛0−𝑏1𝑚0)

𝛼
          (32a)

  

 𝑦′(𝑥) =
(𝑚1𝑥+𝑚0)𝑏0−(𝑛1𝑥+𝑛0)𝑎0

𝛼
 

  =
(𝑏0𝑚1−𝑎0𝑛1)𝑥+(𝑏0𝑚0−𝑎0𝑛0)

𝛼
          (32b) 

Numerical Example 2 

 𝑦′ + 3𝑦 = 6𝑥 + 2 

 −3𝑦′ + 4𝑦 = 8𝑥 − 6 

𝑎1 = 1, 𝑎0 = 3, 𝑏1 = −3, 𝑏0 = 4, 𝑚1 = 6, 𝑚0 = 2, 𝑛1 = 8 , 𝑛0 = −6 

Condition 1:  𝑏0𝑚1 − 𝑎0𝑛1 = (4)(6) − (3)(8) = 0  => Satisfied 

Condition 2: 𝑎1𝑛1 − 𝑏1𝑚1 = 𝑏0𝑚0 − 𝑎0𝑛0 

  (1)(8) − (−3)(6) = (4)(2) − (3)(−6) 

   8 + 18 = 8 + 18   => Satisfied 

 𝛼 = (1)(4) − (3)(−3) = 13 

 𝑦(𝑥) =
[(1)(8)−(−3)(6)]𝑥+[(1)(−6)−(−3)(2)]

13
 

 = 2𝑥 

 𝑦′(𝑥) =
[(4)(6)−(3)(8)]𝑥+[(4)(2)−(3)(−6)]

13
 

 = 2 

Check: 

Eq 1: 𝑦′ + 3𝑦 = 2 + 3(2𝑥) = 6𝑥 + 2  √ 

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Eq 2: −3(2) + 4(2𝑥) = 8𝑥 − 6    √ 

 

4.3 Case 3: 𝒇(𝒙) 𝐚𝐧𝐝 𝒈(𝒙) are exponential functions 

  𝑓(𝑥) = 𝑝𝑒𝑘𝑥   and   𝑔(𝑥) = 𝑞𝑒𝑘𝑥   ,   𝑝, 𝑞, 𝑘 = constant.           (33) 

The necessary condition is 

 𝑎1𝑞𝑘𝑒𝑘𝑥 − 𝑏1𝑝𝑘𝑒𝑘𝑥 = 𝑏0𝑝𝑒𝑘𝑥 − 𝑎0𝑞𝑒𝑘𝑥                 (34) 

which simplifies to 

 𝑘(𝑎1𝑞 − 𝑏1𝑝) = (𝑏0𝑝 − 𝑎0𝑞).             (35) 

Using the above theorem, the solution is given by 

 𝑦(𝑥) =
𝑎1𝑞𝑒𝑘𝑥−𝑏1𝑝𝑒𝑘

𝑎1𝑏0−𝑎0𝑏1
=

𝑎1𝑞−𝑏1𝑝

𝛼
∙ 𝑒𝑘𝑥          (36a) 

 𝑦′(𝑥) =
𝑏0𝑝𝑒𝑘𝑥−𝑎0𝑞𝑒𝑘𝑥

𝑎1𝑏0−𝑎0𝑏1
=

𝑏0𝑝−𝑎0𝑞

𝛼
∙ 𝑒𝑘𝑥          (36b) 

 𝛼 = 𝑎1𝑏0 − 𝑎0𝑏1             (36c) 

Numerical Example 3 

 2𝑦′ + 3𝑦 = 3𝑒3𝑥 

 𝑦′ − 2𝑦 =
1

3
𝑒3𝑥 

𝑎1 = 2, 𝑎0 = 3, 𝑏1 = 1, 𝑏0 = −2, 𝑝 = 3, 𝑞 =
1

3
, 𝑘 = 3 

𝑘(𝑎1𝑞 − 𝑏1𝑝) = 3 (
2

3
− 3) = −7 ; 𝑏0𝑝 − 𝑎0𝑞 = −6 − 1 = −7  =>  Condition satisfied 

𝛼 = (2)(−2) − (3)(1) = −7 

 𝑦(𝑥) =
(2)(

1

3
)−(1)(3)

−7
∙ 𝑒3𝑥 =

1

3
∙ 𝑒3𝑥 

 𝑦′(𝑥) =
(−2)(3)−(3)(

1

3
)

−7
∙ 𝑒3𝑥 = 𝑒3𝑥 

Check: 

Eq 1: 2𝑦′ + 3𝑦 = 2(𝑒3𝑥) + 3 (
1

3
𝑒3𝑥) = 3𝑒3𝑥  √ 

Eq 2: 𝑦′ − 2𝑦 = (𝑒3𝑥) − 2 (
1

3
𝑒3𝑥) =

1

3
𝑒3𝑥  √ 

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 These calculations are programmed into an Excel spreadsheet as shown below Figure 2. 

One just have to enter the corresponding constants and coefficients to obtain the correct answer 

instantly. The necessary condition is also checked and the solution is verified automatically. 

 

 

 

4.4 Case4: 𝒇(𝒙) 𝐚𝐧𝐝 𝒈(𝒙) are sine and cosine functions 

  𝑓(𝑥) = 𝑚 sin 𝑥 + 𝑛 cos 𝑥   and   𝑔(𝑥) = 𝑝 sin 𝑥 + 𝑞 cos 𝑥          (37) 

The necessary condition is 

𝑎1(𝑝 cos 𝑥 − 𝑞 sin 𝑥) − 𝑏1(𝑚 cos 𝑥 − 𝑛 sin 𝑥) = 𝑏0(𝑚 sin 𝑥 + 𝑛 cos 𝑥) − 𝑎0(𝑝 sin 𝑥 + 𝑞 cos 𝑥) 

sin 𝑥 (−𝑎1𝑞 + 𝑏1𝑛 − 𝑏0𝑚 + 𝑎0𝑝) + cos 𝑥 (𝑎1𝑝 − 𝑏1𝑚 − 𝑏0𝑛 + 𝑎0𝑞) = 0        (38) 

which gives 

 𝑎1𝑞 − 𝑎0𝑝 = 𝑏1𝑛 − 𝑏0𝑚            (39a) 

 𝑎1𝑝 + 𝑎0𝑞 = 𝑏1𝑚 + 𝑏0𝑛            (39b) 

The solution is given by 

 𝑦(𝑥) =
(𝑎1𝑝−𝑏1𝑚) sin 𝑥+(𝑎1𝑞−𝑏1𝑛) cos 𝑥

𝛼
          (40a) 

 𝑦′(𝑥) =
(𝑏0𝑚−𝑎0𝑝) sin 𝑥+(𝑏0𝑛−𝑎0𝑞) cos 𝑥

𝛼
          (40b) 

  

  

  

 

 

Figure 2: Spread sheet for Numerical Example 3 

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Numerical Example 4

116𝑦′ − 58𝑦 = −7 sin 𝑥 + 55 cos 𝑥

35𝑦′ + 15𝑦 = 11 sin 𝑥 + 12 cos 𝑥

𝑎1 = 116, 𝑎0 = −58, 𝑏1 = 35, 𝑏0 = 15, 𝑚 = −7, 𝑛 = 55, 𝑝 = 11, 𝑞 = 12

𝛼 = 𝑎1𝑏0 − 𝑎0𝑏1 = 3770



 
 

Necessary conditions: 

(1):  𝑎1𝑞 − 𝑎0𝑝 = 𝑏1𝑛 − 𝑏0𝑚 

 (116)(12) − (−58)(11) = (35)(55) − (15)(−7) 

 2030 = 2030   =>  Condition satisfied 

(2):  𝑎1𝑝 + 𝑎0𝑞 = 𝑏1𝑚 + 𝑏0𝑛 

 (116)(11) + (−58)(12) = (35)(−7) + (15)(55) 

 580 = 580   =>  Condition satisfied 

 𝑦(𝑥) =
[(116)(11)−(35)(−7)] sin 𝑥+[(116)(12)−(35)(55)] cos 𝑥

3770
 

  = 0.403 sin 𝑥 − 0.141 cos 𝑥 

 𝑦′(𝑥) =
[(15)(−7)−(−58)(11)] sin 𝑥+[(15)(55)−(−58)(12)] cos 𝑥

3770
 

  = 0.141 sin 𝑥 + 0.403 cos 𝑥 

Check: 

Eq 1: 116𝑦′ − 58𝑦 = (16.356 sin 𝑥 + 46.748 cos 𝑥) − (23.374 sin 𝑥 − 8.178 cos 𝑥)        

   = −7.018 sin 𝑥 + 54.926 cos 𝑥    √ 

Eq 2: 35𝑦′ + 15𝑦 = (4.935 sin 𝑥 + 14.105 cos 𝑥) + (6.045 sin 𝑥 − 2.115 cos 𝑥)        

   = 10.980 sin 𝑥 + 11.990 cos 𝑥    √ 

 

5. Conclusion 

 A set of two non-homogeneous linear first-order simultaneous ordinary differential 

equations with constant coefficients of a certain form are studied. Four different types of the 

forcing functions, namely constants, linear functions, natural exponential functions and 

sinusoidal functions, have been investigated. For each type, an algebraic equation to determine 

the solution and its derivative as well as the required necessary conditions are derived. Moreover, 

these simple algebraic formulae can be easily programmed into a spreadsheet which can 

automatically compute the solution, check the necessary condition and verify the answer.  

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6. Discussion 

 The theorem and formulae derived are applicable only to one special form of non-

homogeneous linear first-order simultaneous ordinary differential equations with constant 

coefficients. This is indeed restrictive. Moreover, this method only works for the stated four cases 

of the forcing function and a stringent necessary condition must be satisfied. However, most of 

the existing methods have their restrictions too and cannot be used to solve any form of 

simultaneous differential equations with different combinations of the independent and 

dependent variables and their derivatives.  

 Despite its limitations, the proposed theorem and the resulting formulae have the 

advantage that they are algebraic and can be easily programmed into a spreadsheet that can 

perform repeated computations with different constants and coefficients effortlessly. The 

spreadsheet can also be programmed to check the necessary condition and verify the solution. 

The investigation employing this technique to different forms of simultaneous ODE and with 

different forcing functions would be an interesting endeavor. 

 

 

  

 

     

 

    

 

    

   

  

   

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References

1. Andrilli, S. & Hecker, D. (2019). Elementary Linear Algebra (5th edn). Harcourt Academic  Press,

  San Diego.

2. Kolman,  B. & Hill,  D. (2007).  Elementary  Linear  Algebra  (9th edn). MacMillan  Publishing

  Company, New York.

3. Boyce,  W.E. (2017).  Elementary  Differential  Equations  and  Boundary  Value  Problems (11th

  edn). Wiley.

4. Edwards,  H.  &  Penney,  D. (2019). Elementary  Differential  Equations with  Boundary  Value

  Problems (6th edn). Pearson.

5. Kohler, W.E. & Johnson L.W. (2017). Elementary Differential Equations with Boundary Value

  Problems (2nd edn). Pearson.



 
 

6. Zill, D.G. (2018). A First Course in Differential Equations with Modeling Applications (11th edn). 

Cengage. 

7. Differential equations: https://en.wikipedia.org/wiki/Differential_equation 

8. Simultaneous DE: https://www.astro.uvic.ca/~tatum/integrals/integrals06.pdf 

9. Decoupling of Systems of differential equations: Decoupling | Differential Equations | 

Mathematics | MIT OpenCourseWare 

10.   Decoupling: ls4.pdf (mit.edu) 

11. Substitution method: https://web.uvic.ca/~kumara/econ251/schap24.pdf 

12. Elimination method: Ch. 4.8 Solving Systems of D.E.s by Elimination (youtube.com) 

13. D-operator: Microsoft Word - int6 sim eqs.doc (uvic.ca) 

14. How to solve Simultaneous Differential Equations using Laplace Transform:  Bing Videos 

15. Recursive calculations: 2 Solving Sets of Equations (求解聯立方程組) (uotechnology.edu.iq) 

16. Solving System of Simultaneous Differential Equations using Matlab:  

https://www.mathworks.com/help/symbolic/solve-a-system-of-differential-equations.html 

 

 

IJO - INTERNATIONAL JOURNAL OF MATHEMATICS                     (ISSN: 2992-4421 ) IJO JOURNALS

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https://en.wikipedia.org/wiki/Differential_equation
https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/resources/decoupling/
https://ocw.mit.edu/courses/18-03sc-differential-equations-fall-2011/resources/decoupling/
https://math.mit.edu/~jorloff/suppnotes/suppnotes03/ls4.pdf
https://web.uvic.ca/~kumara/econ251/schap24.pdf
https://www.youtube.com/watch?v=nwfh7_J0FBw
https://www.astro.uvic.ca/~tatum/integrals/integrals06.pdf
https://www.bing.com/videos/riverview/relatedvideo?q=methods+to+solve+simultaneous+Differential+equations&mid=4DC227894AE2630800E64DC227894AE2630800E6&FORM=VIRE
https://uotechnology.edu.iq/dep-chem-eng/LECTURE%202014-2015/3Y/numerical/Lecture%209%20Solving%20Simulataneous%20Ordinary%20Differential%20%20%20Equations.pdf
https://www.mathworks.com/help/symbolic/solve-a-system-of-differential-equations.html