Linear Algebra and ODE (MAT228) Chapter - IV Second Order Differential Equations Prof. Zakir H. Ahmed Office# 2030 Phone# 2582172 E-mail: zaahmed@imamu.edu.sa 10/19/2022 Prepared by Prof. Zakir H. Ahmed 1 This Chapter Contains the following Topics: 4.2 Homogeneous Linear Equations 4.3 Auxiliary Equations with Complex Roots 4.4 Nonhomogeneous Equations 4.5 The Superposition Principle etc. Revisited 4.6 Variation of Parameters 4.7 Variable-Coefficient Equations Slides are prepared from the following textbook: • Fundamentals of Differential Equations: R. Kent Nagle, Edward B. Saff, Arthur David Snider, 2012, 8th Edition. 10/19/2022 Prepared by Prof. Zakir H. Ahmed 2 4.2 Homogeneous Linear Equations Consider the following linear second-order constantcoefficient differential equation ay’’ + by’ + cy = f(t) (a ≠ 0) ------------- (1) with the special case where the function f(t) is zero: ay’’ + by’ + cy = 0 ----------------------- (2) Equation (2) is called the homogeneous form of equation (1); f(t) is the “nonhomogeneity” in (1). If we substitute y = ert into (2), we obtain ar2ert + brert + cert = 0 => ert (ar2 + br + c) = 0 => ar2 + br + c = 0 -------------------------- (3) So, y = ert is a solution to (2) iff. r satisfies eq. (3), which is called the auxiliary (characteristic) equation. 10/19/2022 Prepared by Prof. Zakir H. Ahmed 3 4.2 Homogeneous Linear Equations Now the auxiliary equation is just a quadratic, and its roots are: If the discriminant, b2 - 4ac > 0, the roots r1 and r2 are real and distinct. If b2 - 4ac = 0, the roots are real and equal. If b2 - 4ac < 0, the roots are complex conjugate numbers. We consider the first two cases. 10/19/2022 Prepared by Prof. Zakir H. Ahmed 4 4.2 Homogeneous Linear Equations Distinct Real Roots: If the auxiliary equation (3) has distinct real roots r1 and r2, y1(t) = er1t and y2(t) = er2t are solutions to (2) and y(t) = c1er1t + c2er2t is a general solution. Repeated Root: If the auxiliary equation (3) has a repeated root r, y1(t) = ert and y2(t) = tert are solutions to (2) and y(t) = c1ert + c2tert is a general solution. Initial-value problem: For a linear DE, we define 2nd order initial-value problem (IVP) as: Solve: ay’’ + by’ + cy = f(t) Subject to: y(t0) = y0, y’(t0) = y1. 10/19/2022 Prepared by Prof. Zakir H. Ahmed 5 4.2 Homogeneous Linear Equations Theorem 4.1 (Existence and Uniqueness: Homogeneous Case): For any real numbers (a ≠ 0), b, c, t0, y0, and y1, there exists a unique solution to the IVP: ay’ + by’ + cy = 0; y(t0) = y0, y’(t0) = y1 -------- (4) The solution is valid for all t in (- ∞, ∞). Linear Independence of Two Functions: A pair of functions y1(t) and y2(t) is said to be linearly independent (LI) on the interval I iff. neither of them is a constant multiple of the other on all of I. We say that y1 and y2 are linearly dependent on I if one of them is a constant multiple of the other on all of I. Theorem 4.2 (Solutions to IVP): If y1(t) and y2(t) are any two solutions to the DE (2) that are LI on (-∞, ∞), then unique constants c1 and c2 can always be found so that c1y1(t) + c2y2(t) satisfies the IV problem (4) on (-∞, ∞). 10/19/2022 Prepared by Prof. Zakir H. Ahmed 6 4.2 Homogeneous Linear Equations Boundary-value problem: For a 2nd order linear DE, we define boundary-value problem (BVP) as: Solve: a2y’’ + a1y’ + a0y = f(t) Subject to: y(a) = y0, y(b) = y1. Example 4.1: Find a pair of solutions to y’’ + 5y’ - 6y = 0 ------------- (i) Solution: The auxiliary equation associated with (i) is r2 + 5r - 6 = (r – 1)(r + 6) = 0 => r1 = 1, r2 = -6 are roots => y1(t) = et and y2(t) = e-6t are solutions to (i). So, y(t) = c1et + c2e-6t is a general solution. 10/19/2022 Prepared by Prof. Zakir H. Ahmed 7 4.2 Homogeneous Linear Equations Example 4.2: Solve the initial value problem y’’ + 2y’ - y = 0; y(0) = 0, y’(0) = -1 ---------- (i) Solution: The auxiliary equation associated with (i) is r2 + 2r - 1 = 0 => r1 = -1 + √2, r2 = -1 - √2 are roots using the quadratic formula => y(t) = c1e(-1 + √2)t + c2e(-1 - √2)t is a solution. Using initial condition, y(0) = c1e0 + c2e0 => 0 = c1 + c2 => c1 = -c2 and y’(0) = (-1 + √2)c1e0 + (-1 - √2)c2e0 => -1 = (-1 + √2)c1 + (-1 - √2)c2. => c1 = -√2/4 and c2 = √2/4. So, y(t) = (-√2/4)e(-1 + √2)t + (√2/4)e(-1 - √2)t is the solution. 10/19/2022 Prepared by Prof. Zakir H. Ahmed 8 4.2 Homogeneous Linear Equations Example 4.3: Solve the initial value problem y’’ + 4y’ + 4y = 0; y(0) = 1, y’(0) = 3 ---------- (i) Solution: The auxiliary equation associated with (i) is r2 + 4r + 4 = (r + 2)2 = 0 => r = -2 is a double root. => y1(t) = e-2t and y2(t) = te-2t are solutions to (i). and y(t) = c1e-2t + c2te-2t is a general solution. Using initial condition, y(0) = c1e0 + c2(0)e0 => 1 = c1 => c1 = 1 -c2 and y’(t) = -2c1e-2t + c2e-2t -2c2te-2t is => y’(0) = -2c1e0 + c2e0 -2c2(0)e0 => 3 = -2(1)(1) + c2(1) => c2 = 5 So, y(t) = e-2t + 5te-2t is the solution. 10/19/2022 Prepared by Prof. Zakir H. Ahmed 9 4.2 Homogeneous Linear Equations Note: Let’s confirm that and y2(t) = te-2t is a solution. y'2(t) = e-2t - 2te-2t y”2(t) = - 2e-2t - 2e-2t + 4te-2t = - 4e-2t + 4te-2t y”2 + 4y’2 + 4y2 = -4e-2t + 4te-2t + 4(e-2t-2te-2t) +4te-2t = 0 Example 4.4: Find a general solution to the problem y’’’ + 3y’’ –y’ - 3y = 0 ---------- (i) Solution: The auxiliary equation associated with (i) is r3 + 3r2 - r - 3 = (r - 1) (r + 1) (r + 3) = 0 => r1 = 1, r2 = -1 and r3 = -3 are the roots. => y1(t) = et, y2(t) = e-t and y3(t) = e-3t are solutions. So, y(t) = c1et + c2e-t + c3e-3t is a general solution. 10/19/2022 Prepared by Prof. Zakir H. Ahmed 10 4.2 Homogeneous Linear Equations Exercise 4.1: Find a general solution to each given DE. Exercise 4.2: Solve the given initial value problems. Exercise 4.3: Determine whether the functions y1 and y2 are linearly dependent on the interval (0, 1). 10/19/2022 Prepared by Prof. Zakir H. Ahmed 11 To Be Continued…… 10/19/2022 Prepared by Prof. Zakir H. Ahmed 12 End Of Chapter IV 10/19/2022 Prepared by Prof. Zakir H. Ahmed 13