Chapter 8: Linear Systems of Equations MATH 374 Lecture 21 Chapter 8: Linear Systems of Equations

8.1: First Order Systems We now look at systems of linear differential equations. One of the main reasons is that any nth order differential equation with n > 1 can be written as a first order system of n equations in n unknown functions.

Example 1 Write y’’ + y’ – 6y = cos x as a first order system. Solution: Let u = y’. Then u’ = y’’ = – y’ + 6y + cos x = – u + 6y + cos x. Therefore y’ = u u’ = – u + 6y + cos x.

Example 2 Repeat with y’’’ + 10y’’ – 3y’ = x + 14 Solution: Let y1 = y Then y1’ = y’ = y2 y2’ = y’’ = y3 y3’ = y’’’ = – 10y’’ + 3y’ + x + 14 = – 10y3 + 3y2 + x + 14 Therefore y1’ = y2 y2’ = y3 y3’ = – 10y3 + 3y2 + x + 14.

System for nth Order Linear Equation In general, for the differential equation: y(n) + a1y(n-1) + a2y(n-2) + … + an-2y’’ + an-1y’ + any = R(x), (1) if we let y1 = y y2 = y’ y3 = y’’ … yn-1 = y(n-2) yn = y(n-1), Then equation (1) is equivalent to the first order system: y1’ = y2 y2’ = y3 yn-1’ = yn yn’ = – a1yn – a2yn-1 – … – an-2y3 – an-1y2 – any1 + R(x). (2)

Note By concentrating on linear equations with constant coefficients, we will be able to use linear algebra to solve systems of differential equations. For the non-constant coefficient case, many of the solution techniques for first order differential equations carry over to first order systems. See, for example, Chapter 6 of Coddington’s Introduction to Ordinary Differential Equations.

8.2: Solution of a First Order System Consider the system u’ = 4u – v v’ = – 4u + 4v (1) One way to solve this system is to use “algebraic” ideas for solving systems of equations, along with the operator notation from Chapter 4 of our class notes.

Solve using “algebraic” ideas u’ = 4u – v v’ = – 4u + 4v (1) Solve using “algebraic” ideas First, rewrite (1), with operator notation, (D – 4)u + v = 0 (2) 4u + (D – 4)v = 0 (3) Next, operate on (2) with (D – 4) on both sides and subtract (3): [(D – 4)2 – 4]u = 0 (4)

Solve using “algebraic” ideas (D – 4)u + v = 0 (2) [(D – 4)2 – 4]u = 0 (4) Solve using “algebraic” ideas Notice that (4) is a linear equation involving only u! The auxiliary equation of (4) is (m-4)2 – 4 = 0 (5) or m2 – 8m + 12 = 0. (6) The roots of (6) are m = 2, 6. Hence, u = c1e2t + c2e6t. (7)

Solve using “algebraic” ideas (D – 4)u + v = 0 (2) u = c1e2t + c2e6t (7) Solve using “algebraic” ideas To find v, put u from (7) into (2) and solve for v: v = – (D – 4)u = – (D – 4)(c1e2t + c2e6t) = – (2c1e2t + 6c2e6t) + 4(c1e2t + c2e6t) Thus, v = 2c1e2t – 2c2e6t. (8) It follows from (7) and (8) that the solution to (1) is: u = c1e2t + c2e6t. v = 2c1e2t – 2c2e6t. (9)

Another way to solve (1) Here is another way to solve the system u’ = 4u – v v’ = – 4u + 4v (1) Since emt always is in solutions of linear equations, guess a solution to (1) of the form: u = c1emt v = c2emt (10) with c1 and c2 constants and m to be determined.

Another way to solve (1) Substituting (10) into (1) yields: u’ = 4u – v v’ = – 4u + 4v (1) u = c1emt v = c2emt (10) Another way to solve (1) Substituting (10) into (1) yields: c1memt = 4c1emt – c2emt c2memt = – 4c1emt + 4c2emt (11) Dividing out emt in (11), we find: (4 – m)c1 – c2 = 0 – 4c1 + (4 – m)c2 = 0 (12) The only way system (12) can have non-trivial solutions for c1 and c2 is if

Another way to solve (1) which implies (4-m)2 – 4 = 0 or (4 – m)c1 – c2 = 0 – 4c1 + (4 – m)c2 = 0 (12) Another way to solve (1) which implies (4-m)2 – 4 = 0 or m2 – 8m +12 = 0 (14) must hold! (See Theorem 4, Determinants Handout from Section 4.4.)

(4 – m)c1 – c2 = 0 – 4c1 + (4 – m)c2 = 0 (12) m2 – 8m +12 = 0 (14) Another way to solve (1) As we saw above, equation (14) has two solutions, m = 2, 6. It follows from (12), that when m = 2, 2c1 – c2 = 0 – 4c1 + 2c2 = 0 which implies that 2c1 = c2. (Since c1 can be any real number, we say that c1 is a free variable.)

Another way to solve (1) Also, from (12), when m = 6, – 2c1 – c2 = 0 – 4c1 + (4 – m)c2 = 0 (12) m2 – 8m +12 = 0 (14) Another way to solve (1) Also, from (12), when m = 6, – 2c1 – c2 = 0 – 4c1 – 2c2 = 0 which implies that – 2c1 = c2.

u = c1emt v = c2emt (10) Another way to solve (1) Therefore, we have two distinct solutions of form (10), one for m = 2 and one for m = 6: u = c1e2t and u = c1e6t v = 2c1e2t v = –2c1e6t for c1 an arbitrary real number. Compare this form of solution to (9) above!

The Key to Solving System (1) Looking at what we’ve done, the key to solving system (1) seems to be equation (14) and the related system of equations (12)! u’ = 4u – v v’ = – 4u + 4v (1) m2 – 8m +12 = 0 (14) (4 – m)c1 – c2 = 0 – 4c1 + (4 – m)c2 = 0 (12) To formalize what we’ve done, it is useful to use vector and matrix notation and review some linear algebra!!