Revisiting Systems Of Linear Differential Equations

 Recall that the most straightforward type of differential equation to solve is a linear, homogeneous equation.  A linear differential equation is one for which the dependent variable and any of its derivatives exhibit no degree higher than the first.  The general form is described by:

a _o (x ) +  D ⁿ _x y  + a ₁ (x ) +  D ^n-1 _x y  + ……

+   a _n (x ) y    =   f(x)

The differential equation dy/dx = f(x, y) is said to be homogeneous if there exists a function g of one variable such that: 

f(x, y) = g (y/x)

Example: dy/dx = ln x - ln y + (x + y)/ (x - y)

satisfies the definition. Why?

Well, because we can recast it in the form:

dy/dx = ln (x/y) + (1 + y/x) / (1 - y/x)

so that:  g(w) = - ln w + (1 + w)/ (1 - w)

meeting the formal condition of the definition. 

The point is that any homogeneous differential equation can be reduced to a variables separable equation.

One can also solve systems of linear, homogeneous equations. To define homogeneous in this more complex situation, we consider the system of n homogeneous, linear differential equations  as shown below:

dx1/dt = a₁₁(t)x₁ + a₁₂ (t)x₂ + ..a_1n(t)x_n + f₁(t)
:
:
dx2/dt = a₂₁ (t)x₁ + a₂₂ (t)x₂ + ..a_2n (t)x_n + f₂(t)
:
:
dxn/dt = a_n1 (t)x₁ + a_n2 (t)x₂ + ..a_nn(t) x_n + f_n(t)

Where the coefficients are identified as the a_ij and the f_i are functions continuous on a common interval, I.

 When f(t) = 0, and i = 0, 1, 2.......n, this system is said to be homogeneous. Otherwise, it is called non-homogeneous.

The key to solving such a system is that one can use a matrix form to do so and in particular "the fundamental matrix".

If then, X1, X2....X3 is a fundamental set of solutions of the homogenous system:

X' = AX

on an interval, I, then its general solution on the interval is:

X = c1X₁ + c2 X₂ + .......+ c2 X_n


= c1[x_i1_j]  + c2 [x_i2_j]  + ...... cn [x_in_j]

Where the brackets enfold row vectors

Now, let:

X1 = [x_i1_j] and X2 =[x_i2_j] and Xn = [x_in_j]

Be a fundamental set of n solution vectors of the homogeneous system X' = AX on an interval I. Then the matrix:  M(t) =

(x₁₁.....x₁₂........x_1n)

(x₂₁.....x₂₂........x_2n)

(x_n1.....x_n2........x_nn)

is said to be a fundamental matrix of the system on the interval. We want to proceed by solving the linear, homogeneous system:

dx/dt = 2x + 3y

dy/dt = 2x + y

First form a determinant (matrix) from the coefficients: (2, 3) for the top, and (2, 1) for the bottom. Thus:  A =

(2 .....3)
(2......1)

Then, it must be true from the properties of determinants that: (A - lI) D =

[(2 - l)......3]   [k1]

[2 .......(1 -l)]  [k2]

Note how we allow l to be subtracted from the first element in the upper left, and from the last element in the lower right). Cross-multiplying and using matrix properties we obtain the characteristic equation:

(l² -3l +2) – 6 = 0  or    l² -3l - 4 = 0

So that:  (l - 4) (l + 1) = 0,    where l1 = -1 and l2 = 4

You then need to find a vector that solves the equation:

(A -  lI) D = 0

The second eigenvalue is l2 = 4 so you will  repeat the process again to obtain the equation to be solved:

(A - l2 I)D = (A - (4)I) D = (A + I)D = 0 =

[-2.....3] [k1]

[2 ...-.3] [k2]

 

Or: -2k1 + 3k2 = 0 and 2 k1  - 3k2 = 0

For which the values of k1 and k2 can be used to obtain a linearly independent solution.  Here we find our first eigenvector is: k1 =

[1]
[-1]

k2 =

[3]
[2]

Therefore, the first linearly independent solution for the system is:

X1  = k1 exp (-t) 

X2 =  k2 exp(4t)   And we can write the general solution in the form:

Consider the system of 2 linear differential equations:

dx1/ dt = x1 + x2

dx2/dt = 4x1 + x2

  The first step is to form a matrix from the coefficients, which we see are (1, 1) for the top, and (4, 1) for the bottom. Thus:   A =

(1 .....1)

(4......1)

Then, it must be true from the properties of determinants that:

(A - l) D =

[(1 - l).......1] [k1]

[4 ..... .(1 -l)] [k2]

Again, we allow l  to be subtracted from the first element in the upper left, and from the last element in the lower right). Cross-multiplying and using matrix properties we obtain the characteristic equation:

l² - 2l - 3 = 0

where l1 = 3 and l2 = -1 We need to first find a vector that solves the equation:

 (A - l1 I)D = 0

In the first instance, we substitute the first eigenvalue, l= 3, into the matrix:

(A - 3I) D = 0 =

[-2.....1] [k1]

[4 ...-2] [k2]


Therefore:


-2k1 + k2 = 0, and  4k1 – 2k2 = 0,  So k1 = ½ and k2 = 1

Then our first eigenvector is: K1 =

[½]
[1]


Therefore, the first linearly independent solution for the system is:

X1 = K1 exp (l1 t) = K1 exp (3t) 

The second eigenvalue was l2 = -1 so we repeat the process again to obtain the equation to be solved:

(A - l2 I)D = (A - (-1)I) D = (A + I)D

Then, (A + I) D = 0 =


[2.....1] [k1]

[4 ….2] [k2]


Or: 2k1 + k2 = 0 and  4 k1 + 2k2 = 0

So that: k1 = 1, then k2 = - 2

The second eigenvector is then: K2 =


[1]
[-2]

 
So another linearly independent solution is:

X2 = K2 exp (-t)

Then add the two solutions, to obtain: 


X = X1 + X2 = K1 exp (3t) + K2 exp (-t)

With, of course, the 2 column vectors (as computed above) substituted in for K1, K2. This serves as a general approach for solving all such systems.
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Problems for the Math Maven:

1) Obtain the general solution(s) for the system:

dx/dt = 2x + y

 dy/dt = 2x + 3y

2) Obtain the general solution(s) for the system:

dx1/ dt = 3x1 - 4x2

dx2/dt = x1 - x2