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IOSR Journal of Mathematics (IOSR-JM)
e-ISSN: 2278-5728. Volume 5, Issue 3 (Jan. - Feb. 2013), PP 12-17
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Solution of Differential Equations using Exponential of a Matrix
Jervin Zen Lobo 1, Terence Johnson 2
1
(Basic Sciences & Humanities Department, Agnel Institute of Technology and Design, Goa University, India)
2
(Computer Engineering Department, Agnel Institute of Technology and Design, Goa University, India)
Abstract : Our main purpose in this project is to help reader find a clear and glaring relationship between
linear algebra and differential equations, such that the applications of the former may solve the system of the
latter using exponential of a matrix. Applications to linear differential equations on account of eigen values and
eigenvectors, diagonalization of n-square matrix using computation of an exponential of a matrix using results
and ideas from elementary studies form the core study of our project.
Keywords : matrix,fundamental matrix, ordinary differential equations, systems of ordinary
differential equations, eigenvalues and eigenvectors of a matrix, diagonalisation of a matrix, nilpotent
matrix, exponential of a matrix
I.
Introduction
The study of Ordinary Differential Equation plays an important role in our life. Some applications include study
of growth of microorganisms, population, decay of radiation, etc. Ordinary Differential equations is also used in
medicine. Solving a first order Ordinary Differential Equation of first degree could be elementary as we have
many ways of doing so – the Ordinary Differential Equation could be linear, homogenous; or we could solve it
finding suitable integrating factor to make it exact, etc. In solving a second order non-homogenous differential
equation, we have many methods namely: method of undetermined coefficient also called method of judicial
guessing, method of variation of parameters, Inverse D-opertor method, etc. The homogenous part can well
easily be solved by finding the roots of the auxiliary equation. However, as the order of the Ordinary
Differential Equation goes higher, it becomes more tedious to solve the homogenous/non-homogenous part. In
such cases, we reduce the nth order Ordinary Differential Equation into a system of n first order Linear
Differential Equation.
II.
Main Idea Of The Proposed Solution
In this paper we propose to solve a system of first order linear differential equations using techniques of linear
algebra. The innovative part of our paper is the use of exponential of a matrix to find the required solution of the
system.
A2 A3
We know for any matrix A, e  I  A 

 ...
2! 3!
A
If A is nilpotent of degree n, the infinite sum above terminates at some point. In our paper, we use exponential
of matrix in following manner:
Step 1–Write the given system in matrix form. Call the state transition matrix as A.
Step 2 – Find eigenvalues and eigenvectors of A. Form the matrix P whose columns are eigenvectors of A.
Step 3 – Compute P-1AP which will be diagonal. For example P-1AP = diag(11).
Step 4 – Write A = P diag (11) p-1. Hence tA = Pdiag(t1t1t) p-1. Finally etA = Pdiag (et,etet) p-1
Step 5 – Using result that general solution of x1 = Ax is x(t) = etA C. The solution is etA. C; etA being obtained
from step 4
III.
Solution of Differential Equations using Exponential of a Matrix
Theorem: A matrix solution ‘(t)’ of ’=A (t)  is a fundamental matrix of x’=A (t) x iff w (t)  0 for t ϵ
(r1,r2).
Proof: Let (t) be a fundamental matrix of x’ = A (t) x with column vectors 1,2……n.
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Solution of Differential Equations using Exponential of a Matrix
n
Let (t) be a solution of x’=A(t) x. then
 (t )   c j j (t )
for some constants c1,c2,..... cn not all zero.
j 1
By uniqueness of a solution cj’s are all unique.
If c=(c1,c2,…….cn) then (t)=(t)c
For any t0 ϵ (r1,r2) above relation represents n-linear algebraic equations in the n-unknowns c1,c2,………cn and
has unique solution.
 Det (t)0W(t)  0  r1 < t <r2
Let W(t)  0  t ϵ(r1,r2)
1,2……n are linearly independent.
(t) is a fundamental matrix.
Theorem: The general solution of x’=Ax is x(t) = etA c where ‘c’ is an arbitrary constant vector. The solution of
x’=Ax with initial condition ‘x(t0)=x0’ is given by x(t) = e(t-t0)A x0.
Proof: Let x(t) be any solution of x’ =Ax
Define u(t) =e(-tA) x(t) = 0
 u’(t) = e-tA x’(t)- A e(-tA) x(t) = 0
 u (t) = c.
i.e. x(t) = etA c.
Now, x(t0) = et0A c
x0 = et0A c
c = e-t0A x0
 x(t) = etA e-t0A x0
(t--t )A
=e
0
x0
(t)= etA is the fundamental matrix of x’ = Ax, as
’ (t) =
d tA
(e )
dt
= AetA
= A  (t)
Variation of Constant Formula:
Any solution of x’(t) = A(t)x(t) + B(t) such that x(t0) = x0 is given by,
x(t) = (t ) x0   ( s) B( s)ds where ‘ (t ) ’ is a fundamental matrix of
1
x’(t) = A(t)x(t) such that
(t0 ) = I.
Proof: Let x(t) = (t ) C(t), where C(t) is an unknown vector on (r1 , r2 ) .
 x(t0) = (t0 ) C(t 0 )
 C (t0 )  x0 .
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Solution of Differential Equations using Exponential of a Matrix
 x(t )  (t )C(t )  A(t )(t )C (t )
i.e. A(t ) x(t )  B(t )  (t )C(t )  A(t )(t )C(t )
 (t )C(t )  A(t ) x(t )
i.e.B(t )  (t )C(t ).
 C(t )   1 (t ) B(t ).
t
t
t0
t0
  C ( s)ds    1 ( s) B( s)ds
t
i.e.C (t )  C (t0 )    1 ( s) B( s)ds.
t0
t
 C (t )  x0    1 ( s) B( s)ds
t0
t
 x(t )  (t ) x0   (t ) 1 ( s) B( s)ds.
t0
IV.
Experimental Results
Solve the following system of linear first order O.E.E.:
dx
 x  y  4z
dt
dy
 3x  2 y  z
dt
dz
 2x  y  z
dt
Soln: The given system can be put in the form.
1  1 4 
X '  AX , where A  3 2  1
2 1  1
The characteristic polynomial of matrix ‘A’ is
1   1
4
P ( )  A  I  3
2
1
2
1
1  
= (1-)(-3)( +2)
 The eigenvalues of ‘A’ are 1=1, 2=3 and 3=-2
Put 1 = 1 in (A-1) V=0 we get,
0  1 4   v1  0
 3 1  1  v   0 

 2   
2 1  2 v3  0
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Solution of Differential Equations using Exponential of a Matrix
R1R3
2 1  2  v1  0
i.e. 3 1  1 v   0

 2   
0  1 4  v3  0
R1→R1/2
1 1 / 2  1  v1  0
i.e. 3 1  1 v   0

 2   
0  1 4  v3  0
R2→R2-3R1
1 1 / 2  1  v1  0
i.e. 0  1 / 2 2  v   0

 2   
0  1
4  v3  0
R2→R2-3R1
1 1 / 2  1  v1  0
i.e. 3 1  1 v   0

 2   
0  1 4  v3  0
R2→R2-3R1
1 1 / 2  1  v1  0
0  1 / 2 2  v   0 
 2   
i.e. 
0  1
4  v3  0
1
1
R1  R1  R2 , R3  R3  R 2
2
2
1 0 1   v1  0
i.e. 0 1  4 v   0

 2   
0 0 0  v3  0
v1+v3 = 0 and v2-4v3=0
Choose v3 = 1v1 = -1, v2=4.
 1
 
 V1  4 is an eigen vector of ‘A’ corresponding to the eigen value 1=1
 
 1 
1 
 1


 
Similarly, V2  2 is an eigen vector of ‘A’ corresponding to the eigen value 2 = 3 and V3  1 is an
 
 
1 
 1 
eigen vector of ‘A’ corresponding to the eigen value 3 = -2.
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Solution of Differential Equations using Exponential of a Matrix
 1 1  1
  1 2  3
1


1
Let P = 4 2 1 Then, P 
3
0
3 



6
 1 1 1 
 2  2 6 
  1 2  3 1  1 4   1 1  1
1
Now, P AP =
3
0
3  3 2  1  4 2 1 

6
 2  2 6  2 1  1  1 1 1 
-1
1 0 0 

= 0 3
0  = diag (1, 3, -2).

0 0  2
 P-1 AP = diag (1, 3, -2)
 A = P diag (1, 3, -2) P-1
tA = P diag (t, 3t, -2t) P-1
 etA = P diag (et, e3t, e-2t) P-1
t
 1 1  1 e
1

  4 2 1   0
6
 1 1 1   0
 et  3e3t  2e 2t
1
 4e t  6e3t  2e 2t
6
  et  3e3t  2e 2t

0
e 3t
0
   1 2  3

0
3 
 3
e 2t   2  2 6 
0
0
 2et  2e 2t
8et  2e 2t
2et  2e 2t
3et  3e3t  6e 2t 

 12et  6e3t  6e 2t 
 3et  3e3t  6e 2t 
 the general solution of the given system is,
X= etA C.
 et  3e3t  2e 2t
 x
1
 t
 
3t
2 t
i.e. y  4e  6e  2e
  6
  et  3e3t  2e 2t
 z 

 2et  2e 2t
8et  2e 2t
2et  2e 2t
3et  3e3t  6e 2t   c1 

 12et  6e3t  6e 2t  c2 
 3et  3e3t  6e 2t  c3 
where c1, c2, c3 are arbitrary constants.
IV.
Conclusion
Mathematics is one of those few fields of study in which a given problem can be tackled using more
than one approach. In our paper, we have solved systems of differential equations using techniques of linear
algebra by exponential of matrix method. However, there are certain limitations, one of which is the inability to
find the exponential of our matrix if the matrix is non-diagonalizable or the given matrix is not nilpotent or if the
given matrix possesses complex eigen values.
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Solution of Differential Equations using Exponential of a Matrix
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