(I) Method of Elimination

The most elementary approach to linear systems of differential equations involves the elimination of dependent variables by appropriately combining pairs of equations. The object of this procedure is to eliminate dependent variables in succession until there remains only a single equation containing only one dependent variable.

The method of elimination for linear differential systems is similar to the solution of a linear system of algebraic equations by a process of eliminating the unknowns one at a time until only a single equation with a single unknown remains. We consider systems of two equations here; larger systems are discussed in Chapter 5.

Elimination via Successive Differentiation

Example 1

Find the particular solution of the system with \( x(0) = 2 \) and \( y(0) = -1 \):

(1) \( x' = 4x - 3y \)
(2) \( y' = 6x - 7y \)

Solution:

First, isolate \( y \) from equation (1): \( 3y = 4x - x' \).
Differentiate equation (1) with respect to \( t \): \( x'' = 4x' - 3y' \).
Substitute \( y' \) from equation (2) and substitute the isolated expression for \( 3y \):
\( x'' = 4x' - 3(6x - 7y) = 4x' - 18x + 7(3y) \)
\( x'' = 4x' - 18x + 7(4x - x') = 4x' - 18x + 28x - 7x' \).
This simplifies to: \( x'' + 3x' - 10x = 0 \).

The characteristic equation is \( r^2 + 3r - 10 = (r+5)(r-2) = 0 \), yielding roots \( r = 2, -5 \).
General solution for \( x(t) \): \( x(t) = C_1 e^{2t} + C_2 e^{-5t} \).
Using \( y = \frac{4x - x'}{3} \): \( y(t) = \frac{2}{3}C_1 e^{2t} + 3C_2 e^{-5t} \).

Apply initial conditions:
\( x(0) = C_1 + C_2 = 2 \)
\( y(0) = \frac{2}{3}C_1 + 3C_2 = -1 \).
Solving this system yields \( C_1 = 3 \) and \( C_2 = -1 \).

Particular Solution:
\( x(t) = 3e^{2t} - e^{-5t} \)
\( y(t) = 2e^{2t} - 3e^{-5t} \)

Example 2

Solve the following second-order system:

(1) \( x'' = 6x + 2y \)
(2) \( y'' = 3x + 7y \)

Solution:

Differentiate equation (1) twice: \( x^{(4)} = 6x'' + 2y'' \).
Substitute \( y'' \) from (2): \( x^{(4)} = 6x'' + 2(3x + 7y) = 6x'' + 6x + 7(2y) \).
From (1), we have \( 2y = x'' - 6x \). Substituting this in:
\( x^{(4)} = 6x'' + 6x + 7(x'' - 6x) = 13x'' - 36x \).
Standard form: \( x^{(4)} - 13x'' + 36x = 0 \).

Characteristic equation: \( r^4 - 13r^2 + 36 = (r^2 - 4)(r^2 - 9) = 0 \implies r = \pm 2, \pm 3 \).
General solution for \( x(t) \): \( x(t) = C_1 e^{2t} + C_2 e^{-2t} + C_3 e^{3t} + C_4 e^{-3t} \).
Using \( y = \frac{1}{2}x'' - 3x \):
\( y(t) = \frac{1}{2}(4C_1 e^{2t} + 4C_2 e^{-2t} + 9C_3 e^{3t} + 9C_4 e^{-3t}) - 3(C_1 e^{2t} + C_2 e^{-2t} + C_3 e^{3t} + C_4 e^{-3t}) \).
\( y(t) = -C_1 e^{2t} - C_2 e^{-2t} + \frac{3}{2}C_3 e^{3t} + \frac{3}{2}C_4 e^{-3t} \).

(II) Polynomial Differential Operators

We define the operator \( D = \frac{d}{dt} \) and \( D^2 = \frac{d^2}{dt^2} \). A polynomial differential operator \( L \) can be written as \( L = a_n D^n + \dots + a_0 \). These operators are commutative: \( L_1 L_2[x] = L_2 L_1[x] \).

A linear differential system with constant coefficients can be written as:

\( L_1 x + L_2 y = f_1(t) \)
\( L_3 x + L_4 y = f_2(t) \)

Using Cramer's Rule, the operator determinant is \( \Delta = L_1 L_4 - L_2 L_3 \):

\( \Delta x = \begin{vmatrix} f_1 & L_2 \\ f_2 & L_4 \end{vmatrix} \quad \text{and} \quad \Delta y = \begin{vmatrix} L_1 & f_1 \\ L_3 & f_2 \end{vmatrix} \)
Example 3: Hidden Relations

Find the general solution of:

\( (D - 4)x + 3y = 0 \)
\( -6x + (D + 7)y = 0 \)

Solution:

Operator determinant: \( \begin{vmatrix} D-4 & 3 \\ -6 & D+7 \end{vmatrix} = (D-4)(D+7) + 18 = D^2 + 3D - 10 \).
Roots are \( r = 2, -5 \). The general forms are:
\( x(t) = a_1 e^{2t} + a_2 e^{-5t} \)
\( y(t) = b_1 e^{2t} + b_2 e^{-5t} \)

Substitute these into the first equation \( x' - 4x + 3y = 0 \):
\( (2a_1 e^{2t} - 5a_2 e^{-5t}) - 4(a_1 e^{2t} + a_2 e^{-5t}) + 3(b_1 e^{2t} + b_2 e^{-5t}) = 0 \).
Equating coefficients of \( e^{2t} \): \( -2a_1 + 3b_1 = 0 \implies b_1 = \frac{2}{3}a_1 \).
Equating coefficients of \( e^{-5t} \): \( -9a_2 + 3b_2 = 0 \implies b_2 = 3a_2 \).
Result: \( x(t) = a_1 e^{2t} + a_2 e^{-5t} \); \( y(t) = \frac{2}{3} a_1 e^{2t} + 3 a_2 e^{-5t} \).

Remark: When applying the polynomial differential operator method, it is essential to determine the hidden relations between the arbitrary constants \(a_1\), \(a_2\) and \(b_1\), \(b_2\).
Example 4: Nonhomogeneous System

Solve the nonhomogeneous system:

\( (-D - 1)x + (2D - 3)y = e^t \)
\( (3D - 1)x + (-4D + 15)y = e^{-t} \)

Solution:

Step 1: Solve for \( x(t) \).
\( \Delta = (-D-1)(-4D+15) - (2D-3)(3D-1) = -2D^2 - 18 \).
\( (-2D^2 - 18)x = \begin{vmatrix} e^t & 2D-3 \\ e^{-t} & -4D+15 \end{vmatrix} = (-4D + 15)e^t - (2D - 3)e^{-t} = 11e^t + 5e^{-t} \).
Homogeneous roots: \( -2r^2 - 18 = 0 \implies r = \pm 3i \). \( x_c = a_1 \cos(3t) + a_2 \sin(3t) \).
Particular solution guess \( x_p = Ae^t + Be^{-t} \) yields \( A = -11/20, B = -1/4 \).
\( x(t) = a_1 \cos(3t) + a_2 \sin(3t) - \frac{11}{20}e^t - \frac{1}{4}e^{-t} \).

Step 2: Solve for \( y(t) \).
\( (-2D^2 - 18)y = \begin{vmatrix} -D-1 & e^t \\ 3D-1 & e^{-t} \end{vmatrix} = (-D-1)e^{-t} - (3D-1)e^t = -2e^t \).
\( y_c = b_1 \cos(3t) + b_2 \sin(3t) \) and \( y_p = \frac{1}{10}e^t \).
\( y(t) = b_1 \cos(3t) + b_2 \sin(3t) + \frac{1}{10}e^t \).

Step 3: Hidden Relations.
Substitute back to find relations between \( a_1, a_2 \) and \( b_1, b_2 \):
For \( \sin(3t) \): \( 3a_1 - 6b_1 = a_2 + 3b_2 \) and \( 6b_1 + 3b_2 = 3a_1 - a_2 \).
For \( \cos(3t) \): \( a_1 + 3b_1 = -3a_2 + 6b_2 \) and \( 3b_1 - 6b_2 = -a_1 - 3a_2 \).
Solving gives \( b_1 = \frac{1}{2}(a_1 - a_2) \) and \( b_2 = \frac{1}{2}(a_1 + a_2) \).

Final Form: \( y(t) = \frac{1}{2}(a_1 - a_2) \cos(3t) + \frac{1}{2}(a_1 + a_2) \sin(3t) + \frac{1}{10}e^t \).