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# Notes-HigherOrderLinEq.pdf

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School
Pennsylvania State University
Department
Mathematics
Course
MATH 251
Professor
Unknown
Semester
Spring

Description
Higher Order Linear Equations with Constant Coefficients The solutions of linear differential equations with constant coefficients of the third order or higher can be found in similar ways as the solutions of second order linear equations. For an n▯th order homogeneous linear equation with constant coefficients: (n) (n−1) any + a n−1 y + … + a y″ 2 a y′ + 1 y = 0,0 a n 0. It has a general solution of the form y = C y1 1C y + 2 2 C n−1 n−1 + C n n where y 1 y 2 … , y n−1, n are any n linearly independent solutions of the equation. (Thus, they form a set of fundamental solutions of the differential equation.) The linear independence of those solutions can be determined by their Wronskian, i.e., W(y ,1y 2 … , y n−1 yn)(t) ≠ 0. Note 1: In order to determine the n unknown coefficients C i each n▯th order equation requires a set of n initial condi(n−1) in an(n−1)ial value problem: y(t0) = y0, y′0t ) = 0′ , y0(t ) =0y″ , and y (0 ) = y 0. Note 2: The Wronskian W(y , y ,1… ,2y n−1 yn)(t) is defined to be the determinant of the following n × n matrix  y1 y2 .. .. yn    y'1 y' 2 .. .. y'n    y" 1 y" 2 .. .. y" n    .  : : :  (n 1) (n−1) (n−1)  y1 y 2 .. .. yn  © 2008 Zachary S Tseng B▯4 ▯ 1 Such a set of linearly independent solutions, and therefore, a general solution of the equation, can be found by first solving the differential equation’s characteristic equation: a r + a r n−1 + … + a r + a r + a = 0. n n−1 2 1 0 This is a polynomial equation of degree n, therefore, it has n real and/or complex roots (not necessarily distinct). Those necessary n linearly independent solutions can then be found using the four rules below. rt (i). If r is a distinct real root, then y = e is a solution. (ii). If r = λ ± ▯i are distinct complex conjugate roots, then y = e cos ▯t and y = e sin ▯t are solutions. (iii). If r is a real root appearing k times, then y = e rt, y = te rt, 2 rt k−1 rt y = t e , … , and y = t e are all solutions. (iv). If r = λ ± ▯i are complex conjugate roots each appears k times, then y = e cos ▯t , y = e sin ▯t , λ t λ t y = te2coλ tt , y = te 2inλ t , y = t e cos ▯t , y = t e sin ▯t , : : k−1 : λ t : k−1 λ t y = t e cos ▯t , and y = t e sin ▯t , are all solutions. © 2008 Zachary S Tseng B▯4 ▯ 2 (4) Example: y − y = 0 4 2 The characteristic equation is r − 1 = (r + 1)(r + 1)(r − 1) = 0, which has roots r = 1, −1, i, −i. Hence, the general solution is y = C e + C e −t + C cost + C sint. 1 2 3 4 (5) (4) (3) Example: y − 3y + 3y − y″ = 0 The characteristic equation is r − 3r + 3r − r = r (r − 1) = 0, 3 which has roots r = 0 (a double root), and 1 (a triple root). Hence, the general solution is y = C e + C te + C e + C te + C t e t 2 t 1 2 3 4 5 t t 2 t = C +1C t +2C e + C3te + C t4e . 5 Example: y(4) + 4y (3)+ 8y″ + 8y′ + 4y = 0 4 3 2 2 2 The characteristic equation is r + 4r + 8r + 8r + 4 = (r + 2r + 2) = 0, which has roots r = −1 ± i (repeated). Hence, the general solution is − t − t − t − t y = C e1 cost + C e 2 sin t + C t3 cost + C te4 sin t. © 2008 Zachary S Tseng B▯4 ▯ 3 Example: What is a 4th order homogeneous linear equation whose general solution is t 2t 3t 4t y = C 1 + C e + 2 e + C e 3 4 The solution implies that r = 1, 2, 3, and 4 are the four roots of the characteristic equation. Therefore, r − 1, r − 2, r − 3, and r − 4 are its factors. Consequently, the characteristic equation is (r − 1)(r − 2)(r − 3)(r − 4) = 0 4 3 2 r − 10r + 35r − 50r + 24 = 0 Hence, an equation is (4) (3) y − 10y + 35y″ − 50y′ + 24y = 0. Note: The above answer is not unique. Every nonzero constant multiple of the above equation also has the same general solution. However, the indicated equation is the only equation in the standard form that has the given general solution. © 2008 Zachary S Tseng
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