Solutions to systems of simultaneous linear differential equations with constant coefficients

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We shall now consider systems of simultaneous linear differential equations which contain a single independent variable and two or more dependent variables. In general, the number of equations will be equal to the number of dependent variables i.e. if there are n dependent variables there will be n equations.

Examples of systems

Example 1. The following two-equation system where x and y are dependent variables and t is the independent variable:

or, equivalently,

2(D - 2)x + (D - 1)y = e^t

(D + 3)x + y = 0

where D = d/dt.

Example 2. The following three-equation system where x, y and z are dependent variables and t is the independent variable:

or, equivalently,

Dx + (D + 1)y = 1

(D + 2)x - (D - 1)z = 1

(D + 1)y + (D + 2)z = 0

where D = d/dt.

Each equation in a system will be assumed to have constant coefficients and be of the general form

f(D)x + g(D)y + ....... + h(D)u = G(t)

where x, y, ...... , u are the dependent variables, t is the independent variable.

A system of n equations in the n dependent variables x, y, ..... , u and independent variable t will have a solution consisting of n functions

x = x(t)

y = y(t)

.............

u = u(t)

The method used to solve a system of n equations in n variables is analogous to the procedure used to solve a system of n linear equations in n unknowns in algebra. In algebra, we solve a system of n equations in n unknowns by eliminating unknowns between equations until we obtain an equation containing a single unknown, from which we deduce the value of the unknown. Then we substitute the value of that unknown into other equations to obtain the values of other unknowns. In solving systems of linear differential equations we go through the same type process to obtain an equation containing a single dependent variable. The equation in this single dependent variable will be a linear differential equation with constant coefficients. We then solve this equation, using methods for solving such equations, to obtain an expression for that dependent variable. We then substitute the expression for that variable into another equation to obtain an expression for another variable. As with the algebraic problem, we can also employ determinants.

Theorem. The number of constants in the general solution of a system of equations must equal the sum of the orders of the equations.

Problem 1. Solve the linear system

1) y" - y + 5v' = x

2) 2y' - v" + 4v = 2

Solution.

Method 1.

Step 1. Put the equations into operator form.

3) (D² - 1)y + 5Dv = x

4) 2Dy - (D² - 4)v = 2

Step 2. Obtain an equation in v alone by eliminating y.

Multiplying 3) through by 2D and 4) through by (D² - 1) and then subtracting one from the other we obtain

[10D² + (D² - 1)(D² - 4)]v = 2Dx - (D² -1)2

5) (D⁴ + 5D² + 4)v = 4

Note. When we speak of “multiplying” by an operator we really mean “operate on” by the operator.

Step 3. Solve 5) for v.

Using methods for solving linear differential equations with constant coefficients we find the solution as

6) v = 1 + a₁ cos x + a₂ sin x + a₃ cos 2x + a₄ sin 2x

Step 4. Obtain an equation in y alone.

We could, if we wished, find an equation in y using the same method as we used in Step 2. However, for the purpose of demonstration, we will find y using determinants. Using Cramer’s Rule on the system of equations 3) and 4) we get

[-(D² - 1)(D² - 4) - 10D²] y = 4x

8) (D⁴ + 5D² + 4) y = - 4x

Note. We must take care to interpret the right member of 7) correctly. It is interpreted as

-(D² - 4)( x) - 5D(2)

Step 5. Solve 8) for y. The solution is

9) y = -x + b₁ cos x + b₂ sin x + b₃ cos 2x + b₄ sin 2x

Now the number of constants in the solution must be four since the number of constants must equal the sum of the orders of the equations in the original system. Since 1) is of order 2 and 2) is of order 2) the sum of the orders is four. Now 6) and 9) must satisfy the original equations rather than just equations 5) and 8) which resulted from the original ones after some manipulations were performed (the technique we used introduced extra constants).

Step 6. Substitute 6) and 9) into equation 1). We get

10) x - 2b₁ cos x - 2b₂ sin x - 5b₃ cos 2x - 5b₄ sin 2x - 5a₁ sin x + 5a₂ cos x - 10a₃ sin 2x + 10a₄ cos 2x x

which is an identity in x.

Step 7. Equate coefficients of corresponding terms in 10) to obtain the final coefficients.

The fact that 10) is an identity in x demands that

- 2b₁ + 5a₂ = 0

- 2b₂ - 5a₁ = 0

- 5b₃ + 10a₄ = 0

- 5b₄ - 10a₃ = 0

Note. If we were to go through steps 6 and 7 for equation 2) we would get equivalent results.

Step 8. Write down the general solution. It is

v = 1 + a₁ cos x + a₂ sin x + a₃ cos 2x + a₄ sin 2x

************************************************

Method 2. We shall now solve the same problem by a different method which is simpler and easier. It is the same as Method 1 up through Step 3, but finds y in a different way.

Step 1. Put the equations into operator form.

3) (D² - 1)y + 5Dv = x

4) 2Dy - (D² - 4)v = 2

Step 2. Obtain an equation in v alone by eliminating y.

Multiplying 3) through by 2D and 4) through by (D² - 1) and then subtracting one from the other we obtain

[10D² + (D² - 1)(D² - 4)]v = 2Dx - (D² -1)2

5) (D⁴ + 5D² + 4)v = 4

Step 3. Solve 5) for v.

Using methods for solving linear differential equations with constant coefficients we find the solution as

6) v = 1 + a₁ cos x + a₂ sin x + a₃ cos 2x + a₄ sin 2x

Step 4. Find an equation giving y in terms of v. We accomplish this by eliminating from the system of 3) and 4) those terms which involve derivatives of y.

Multiplying 3) by 2 and 4) by D we get

7) (2D² - 2)y + 10Dv = 2x

8) 2D²y - (D³ - 4D)v = 0

Subtracting 7) from 8) we get

9) 2y - D³v - 6Dv = -2x

Step 5. Substitute 6) into 10) to obtain the expression for y. It is

Step 6. Write down the general solution. It is

v = 1 + a₁ cos x + a₂ sin x + a₃ cos 2x + a₄ sin 2x

We have given both methods to illustrate the superiority of the second method. The second method avoids introducing extra constants. Thus it is best to use the second method whenever possible.

We will now give another example to make a point. The point to be made is that the use of differential operators is not necessary to solving systems of linear differential equations. They make the work much easier but what can be done with them can also be done without them.

Problem 2. Solve the following system