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 + ..a1n(t)xn
+ f1(t)
:
:
dx2/dt = a21 (t)x1 + a22 (t)x2 +
..a2n (t)xn + f2(t)
:
:
dxn/dt = an1 (t)x1 + an2 (t)x2 +
..ann(t) xn + fn(t)
Where the coefficients are identified as the aij and the fi
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 Xn
= c1[xi1j] + c2 [xi2j]
+ ...... cn [xinj]
Where the brackets enfold row vectors
Now, let:
X1 = [xi1j] and X2 =[xi2j]
and Xn
= [xinj]
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........x1n)
(x21.....x22........x2n)
(xn1.....xn2........xnn)
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)
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 - 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.
---------
dy/dt = 2x + 3y
2) Obtain the general solution(s) for the system:
dx1/ dt = 3x1 - 4x2
dx2/dt = x1 - x2
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