Section 5.1 First-Order Systems & Applications

Section 5.1First-Order Systems & Applications

Suppose x and y are both functions of t. Solve: x′ = 3x – y y′ = 2x + y – et

Why would we consider such things?

Ex. 1 Give the system of diff eqs which describe the following tanks containing brine solution:

Theorem: Consider the following system of diff eqs (all xi, pij, and fiare functions of t ) x1′ = p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1 x2′ = p21x1 + p22x2 + p23x3 + ⋯ + p2nxn + f2 x3′ = p31x1 + p32x2 + p33x3 + ⋯ + p3nxn + f3 : : xn′ = pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn Let J be an open interval containing t = a. Suppose the functions pij and the functions fk are continuous on J. Then the system of differential equations has a unique solution that satisfies the following initial conditions: x1(a) = b1 , x2(a) = b2 , x3(a) = b3 , ......... , xn(a) = bn

If we have a system of higher order diff eqs, we can transform it into a system of first order diff eqs.

Ex. 3 Rewrite the one diff eqx(3) + 3x″ + 2x′ – 5x = sin(2t) as a system of first order diff eqs.

Ex. 4 Rewrite the following system as a system of first order diff eqs. 2x″ = –6x + 2y y″ = 2x – 2y + 40sin(3t)

Section 5.2The Method Of Elimination

Review of solving systems of algebraic equations (2 equations with 2 unknowns) i. Substitution method ii. Elimination method iii. Cramer's rule

Ex. 1 Find the general solution to the following system by using a variant of the substitution method. x′ = 4x – 3y y′ = 6x – 7y

Ex. 2 Find the general solution to the following system by using a variant of the elimination method. x′ = 4x – 3y y′ = 6x – 7y

We shall now use the following notation: L will be a linear operator of the form L = anDn + an–1Dn–1 + an–2Dn–2 + ⋯ + a2D2 + a1D + a0

Ex. 3 Find the general solution to the following general system of diff eqs: L1x + L2y = f1(t) L3x + L4y = f2(t)

Ex. 4 Find the general solution to x′ = 2x + y y′ = 2x + 3y + e5t Solution: (D–2)x – y = 0 –2x + (D–3)y = e5t [(D–2)(D–3) – 2] x = (D – 3)0 + e5t [(D–2)(D–3) – 2] y = (D – 2) e5t + 2(0) (D2–5D+4) x = e5t (D2–5D+4) y = 5e5t – 2e5t (D–4)(D–1) x = e5t (D–4)(D–1) y = 3e5t xc = c1e4t + c2etyc = c3e4t + c4et xp = Ae5t ⇒ xp = (1∕4)e5typ = Be5t ⇒ yp = (3∕4)e5t x = c1e4t + c2et + (1∕4)e5ty = c3e4t + c4et + (3∕4)e5t

Ex. 4 Find the general solution to x′ = 2x + y y′ = 2x + 3y + e5t Solution: x = c1e4t + c2et + (1∕4)e5ty = c3e4t + c4et + (3∕4)e5t Plugging these solutions into the first equation in the initial problem we get: x′ = 2x + y (c1e4t + c2et + (1∕4)e5t)′ = 2(c1e4t + c2et + (1∕4)e5t) + (c3e4t + c4et + (3∕4)e5t) 4c1e4t + c2et + (5/4)e5t = 2c1e4t + 2c2et + (1/2)e5t + c3e4t + c4et + (3/4)e5t 0 = (–2c1 + c3)e4t + (c2 + c4)et –2c1 + c3 = 0 ⇒ c3 = 2c1 c2 + c4 = 0 c4 = –c2

Section 5.3Matrices & Linear Systems

x1′ = 2x1 + x2 x2′ = 2x1 + 3x2

x1′ = 2x1 + x2 ⇒ x2′ = 2x1 + 3x2

Previously we would state: The general solution to the system x1′ = 2x1 + x2 turns out to be: x1 = c1e4t + c2et x2′ = 2x1 + 3x2 x2 = 2c1e4t – c2et

Previously we would state: The general solution to the system x1′ = 2x1 + x2 turns out to be: x1 = c1e4t + c2et x2′ = 2x1 + 3x2 x2 = 2c1e4t – c2et We now would state: The general solution to the system turns out to be:

Previously we would state: The general solution to the system x1′ = 2x1 + x2 turns out to be: x1 = c1e4t + c2et x2′ = 2x1 + 3x2 x2 = 2c1e4t – c2et We now would state: The general solution to the system turns out to be: We shall find that general solutions to these "2x2 homogenous systems" will take this form of (where and are two linearly independent vectors).

x1′ = p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1 x2′ = p21x1 + p22x2 + p23x3 + ⋯ + p2nxn + f2 x3′ = p31x1 + p32x2 + p33x3 + ⋯ + p3nxn + f3 : : xn′ = pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn

x1′ = p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1 x2′ = p21x1 + p22x2 + p23x3 + ⋯ + p2nxn + f2 x3′ = p31x1 + p32x2 + p33x3 + ⋯ + p3nxn + f3 : : xn′ = pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn This system of diff eqs is said to be homogenous if

x1′ = p11x1 + p12x2 + p13x3 + ⋯ + p1nxn + f1 x2′ = p21x1 + p22x2 + p23x3 + ⋯ + p2nxn + f2 x3′ = p31x1 + p32x2 + p33x3 + ⋯ + p3nxn + f3 : : xn′ = pn1x1 + pn2x2 + pn3x3 + ⋯ + pnnxn + fn This system of diff eqs is said to be homogenous if Thus, the matrix equation would be for a homogenous system.

We shall now see many theorems very similar to theorems and definitions we had in chapter 2.

Definition: are said to be linearly independent if the equation only has the solution of c1 = c2 = ⋯ = cn = 0.

Definition: If are solutions to then we define the Wronskian of these solution to be

Theorem: Suppose are solutions to on an interval J where all the pij functions are continuous. (a) If are linearly dependent then W = 0 for every point on the interval J. (b) If are linearly independent then W ≠ 0 for every point on the interval J.

Theorem: Suppose are linearly independent solutions to on an interval J where all the pij functions are continuous. The general solution to is given as:

Theorem: Suppose is a particular solution to the nonhomogenous system and is the general solution to the corresponding homogenous system . Then the general solution to the nonhomogenous system is .

Ex. 1x′1 = x1 + x2 – 2x3 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3 (a) Write this system as a matrix equation.

Ex. 1x′1 = x1 + x2 – 2x3 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3 (b) Verify that the following three vectors are solution vectors:

Ex. 1x′1 = x1 + x2 – 2x3 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3 (c) Verify that these are linearly independent vectors.

Ex. 1x′1 = x1 + x2 – 2x3 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3 (d) Give the general solution for the system.

Ex. 1x′1 = x1 + x2 – 2x3 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3 (e) Give the general solution for x1. Give the general solution for x2. Give the general solution for x3.

Ex. 1x′1 = x1 + x2 – 2x3 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3 (f) Solve the initial value problem: x′1 = x1 + x2 – 2x3x1(0) = 9, x2(0) = –2, x3(0) = 5 x′2 = –x1 + 2x2 + x3 x′3 = x2 – x3

Section 5.4The Eigenvalue Method for Homogeneous Systems

Review of eigenvalues and eigenvectors: Let A be a square matrix. The vector is said to be an eigenvector for the eigenvalue λ if .

Review of eigenvalues and eigenvectors: Let A be a square matrix. The vector is said to be an eigenvector for the eigenvalue λ if . What is the connection between eigenvectors and solutions to systems of diff eqs?

Ex. 1 We previously (in section 5.3) found three linearly independent solution vectors to the system One of these was Verify that this solution is an eigenvector of

Ex. 2 Suppose that (where A, B, C, and k are constants) is a solution vector to the system . Show that this solution vector is an eigenvector of the matrix P.

Theorem: Given a homogenous system , suppose λ is an eigenvalue of P with eigenvector . Then is a nontrivial solution vector to .

Ex. 3 Use eigenvalues/eigenvectors to find the general solution to the following system and write your final solution in scalar form. x′1 = 3x1 + x2 x′2 = 3x1 + 5x2

Note: Suppose where P is an nxn matrix. If there are n distinct real eigenvalues of P then we've got n linearly independent solution vectors. This means we can write down the general solution as (here the λi are the eigenvalues, the are the eigenvectors and the ci are arbitrary constants). If we have less than n distinct eigenvalues, or if some of the eigenvalues are complex then we will run into trouble and need to do something else.

Section 5.1 First-Order Systems & Applications