System of Differential Equations

SYSTEM OF DIFFERENTIAL EQUATIONS

Example 18

Consider the system of differential equations

x'₁=3x₁+x₂+x₃

x'₂ = 2x₁ + 4x₂ + 2x₃

x'₃ = −x₁−x₂+x₃

where, for each i, x_i=x_i(t) is a differential real−valued function of the real variable t. Find the general  solution:

Solution

Let x: R → R³ be the function defined by

The derivative of x denoted by x', is defined by 

 (i.e) x'(t) = Ax

where   is the coefficient matrix of the given system.

Eigenvalues of the matrix A are 2, 2, 4.

The corresponding basis for the eigenvalues are obtained as

The diagonalized matrix of Q is obtained as

 D = Q‒1AQ = 

Substitute A = QDQ^‒1 into x' = Ax to obtain

 x′(t) = (QDQ^‒1)x or Q^‒1x' = DQ^‒1x

Let the function y: R → R³ defined by

 y(t) = Q^‒1x (t). It can be differentiable and y'=Q^‒1x'

The original system can be written as

y' = Dy.  Let  

we can rewrite y' = Dy as follows

By comparing on both sides, the three equations are as follows.

 y'₁ = 2y₁, y'₂=2y₂ and y₃'=4y₃, are independent to each other.

The general solution of these equation is

 y₁(t) = c₁e^2t, y₂(t) = c₂e^2t, y₃(t) = c₃e^4t

where c₁, c₂ and c₃ are arbitrary constants.

Finally,

 yields the general solution, of the original system. This solution can be written as

  The general solution is

 x(t) = e^2tz₁ + e^4tz₂

where

 z₁ ∈ E_λ1, and z₂ ∈ E _λ2

Example 19

Find the general solution of the system of differential equations;

 x₁'=x₁+x₃; x₂=x₂+x₃ & x₃' = 2x³.

Solution:

Let

The derivative of x denoted by x', is defined by 

Let   be the coefficient matrix of the given system.

Then the system as the matrix equation can be written as x¹ = Ax

                      ……………….(1)

The eigenvalues of the matrix A are 1, 1, 2.

The corresponding basis for the eigenvalues are obtained as

Then

Substitute A=QDQ^‒1 in to x'=Ax to obtain x'(t) = (QDQ^‒1)x (or) Q^‒1x' = DQ^‒1x

Let the function y: R→ R³ defined by

  y(t) = Q^‒1x (t) ⇒ y′(t) = Q^‒1x'.

  The original system can be written as y' = Dy.

Then y' = Dy can be written as

This gives y'₁=y₁(t), y'₂=y₂(t), y₃' = 2y₃(t).

The general solution of these equations are y₁(t) = c₁e^t, y₂(t) = c₂e^t & y₃ (1) = c₃ e^2t.

Finally

which yields general solution of the original system.

This solution can be written as

 The general solution is x(t) = e^t[z₁ + z₂ +z₃]

where