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MATH 374 Lecture 21

Learn how to write nth-order differential equations as first-order linear systems, with examples and solution techniques. Explore using algebraic methods and operator notation. Discover various ways to solve first-order systems with constant coefficients using linear algebra.

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MATH 374 Lecture 21

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

  2. 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.

  3. 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.

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

  5. 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)

  6. 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.

  7. 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.

  8. 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)u + v = 0 (2) [(D – 4)2 – 4]u = 0 (4)

  9. (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)

  10. (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)

  11. 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.

  12. 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

  13. (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.)

  14. (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.)

  15. (4 – m)c1 – 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.

  16. 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!

  17. 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!!

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