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 = a_{11}(t)x_{1} + a_{12 }(t)x_{2} + ..a_{1n}(t)x_{n}
+ f_{1}(t)

:

:

dx2/dt = a_{21 }(t)x_{1} + a_{22 }(t)x_{2} +
..a_{2n }(t)x_{n} + f_{2}(t)

:

:

dxn/dt = a_{n1 }(t)x_{1} + a_{n2 }(t)x_{2} +
..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 = c1X**_{1}** + c2 X**_{2}** + .......+ c2 X**_{n}

= c1[x_{i}1_{j}] + c2 [x_{i}2_{j}]
+ ...... cn [x_{i}n_{j}]

Where the brackets enfold *row vectors*

Now, let:

**X1** = [x_{i}1_{j}] and **X2** =[x_{i}2_{j}]
and **Xn**
= [x_{i}n_{j}]

Be a fundamental set of n solution vectors of the homogeneous system

*on an interval I. Then the matrix: M(t) =*

**X' = AX**(x

_{11}.....x

_{12}........x

_{1n})

(x

_{21}.....x

_{22}........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:

(l^{2} -3l +2) – 6 = 0 *or* l^{2} -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:
-2**k1** + 3**k2** = 0 and 2 **k1** - 3**k2 **= 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:

l

^{2}- 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:

-2

**k1**+

**k2**= 0, and 4

**k1**– 2

**k2**= 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: 2

**k1**+

**k2**= 0 and 4

**k1**+ 2

**k2**

**= 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.

---------

__Problems for the Math Maven__: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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