Chapter 4: Higher Order Linear Equations

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1 Chapter 4: Higher Order Linear Equations MATH 351 California State University, Northridge April 7, 2014 MATH 351 (Differential Equations) Ch 4 April 7, / 11

2 Sec. 4.1: General Theory of nth Order Linear Equations In general, an nth order DE Where P 0 (t) d n y dt n + P 1(t) d n 1 y dt n P n 1(t) dy dt + P n(t)y = G(t) (1) P 0,..., P n and G are continuous real-valued functions on some interval I : α < t < β; P 0 is nowhere zero in I. We can define the linear differential operator L of order n by L[y] = P 0 (t) d n y dt n + P 1(t) d n 1 y dt n P n 1(t) dy dt + P n(t)y MATH 351 (Differential Equations) Ch 4 April 7, / 11

3 The nth order linear IVP L[y] = d n y dt n + p 1(t) d n 1 y dt n p n 1(t) dy dt + p n(t)y = g(t) (2) Subject to y(t 0 ) = y 0, y (t 0 ) = y 0,..., y (n 1) (t 0 ) = y (n 1) 0 MATH 351 (Differential Equations) Ch 4 April 7, / 11

4 Existence and Uniqueness Theorem For 2nd order Consider the initial value problem y + p(t)y + q(t)y = g(t), y(t 0 ) = y 0, y (t 0 ) = y 0 (3) where p, q and g are continuous on an open interval I that contains the point t 0. Then there is exactly one solution y = φ(t) of this problem, and the solution exists throughout the interval I. MATH 351 (Differential Equations) Ch 4 April 7, / 11

5 Existence and Uniqueness Theorem For 2nd order Consider the initial value problem y + p(t)y + q(t)y = g(t), y(t 0 ) = y 0, y (t 0 ) = y 0 (3) where p, q and g are continuous on an open interval I that contains the point t 0. Then there is exactly one solution y = φ(t) of this problem, and the solution exists throughout the interval I. For nth order If the functions p 1, p 2,, p n and g are continuous on the open interval I, then there exists exactly one solution y = φ(t) of the nth order linear IVP (on the previous page), where t 0 is any point in I. This solution exists throughout the interval I. MATH 351 (Differential Equations) Ch 4 April 7, / 11

6 General Solution for the Homogeneous Equations For the homogeneous equation L[y] = d n y dt n The general solution where + p1(t) d n 1 y dt n 1 c 1, c 2,, c n are arbitrary constants. dy + + pn 1(t) dt + pn(t)y = 0 y = c 1y 1(t) + c 2y 2(t) + + c ny n(t) y 1, y 2,, y n a fundamental set of solutions (whose Wronskian is nonzero) (Linearly Independent) MATH 351 (Differential Equations) Ch 4 April 7, / 11

7 General Solution for the Homogeneous Equations For the homogeneous equation L[y] = d n y dt n The general solution where + p1(t) d n 1 y dt n 1 c 1, c 2,, c n are arbitrary constants. dy + + pn 1(t) dt + pn(t)y = 0 y = c 1y 1(t) + c 2y 2(t) + + c ny n(t) y 1, y 2,, y n a fundamental set of solutions (whose Wronskian is nonzero) (Linearly Independent) FSSs & Linearly Independent If y 1(t),..., y n(t) is a fundamental set of solutions of L[y] = d n y dt n + p1(t) d n 1 y dt n 1 dy + + pn 1(t) dt + pn(t)y = 0 on an interval I, then y 1(t),..., y n(t) are linearly independent on I. Conversely, if y 1(t),..., y n(t) are linearly independent solutions of the above equation on I, then they form a fundamental set of solutions on I. MATH 351 (Differential Equations) Ch 4 April 7, / 11

8 The Nonhomogeneous Equation Consider the nonhomogeneous equation L[y] = d n y dt n The general solution is where + p1(t) d n 1 y dt n 1 dy + + pn 1(t) + pn(t)y = g(t) (4) dt y = c 1y 1(t) + c 2y 2(t) + + c ny n(t) + Y (t) (5) y 1, y 2,..., y n are the fundamental set of solutions of the corresponding homogeneous equation; Y (t) particular solution of the nonhomogeneous equation MATH 351 (Differential Equations) Ch 4 April 7, / 11

9 The Nonhomogeneous Equation Consider the nonhomogeneous equation L[y] = d n y dt n The general solution is where + p1(t) d n 1 y dt n 1 dy + + pn 1(t) + pn(t)y = g(t) (4) dt y = c 1y 1(t) + c 2y 2(t) + + c ny n(t) + Y (t) (5) y 1, y 2,..., y n are the fundamental set of solutions of the corresponding homogeneous equation; Y (t) particular solution of the nonhomogeneous equation To determine a fundamental set of solutions y 1, y 2,..., y n constant coefficients fairly simple, discuss later. non-constant coefficients numerical method (Ch 8) or series method (Ch 5) To find Y (t) the method of undetermined coefficients and variation of parameters. MATH 351 (Differential Equations) Ch 4 April 7, / 11

10 Sec. 4.2: Homogeneous Equations with Constant Coefficients Consider the nth order linear homogeneous differential equation d n y L[y] = a 0 dt n + a d n 1 y 1 dt n a dy n 1 dt + a ny = 0 (6) where a 0, a 1,..., a n are real constants, and a 0 0 Use the test function y = e rt, we obtain i.e., L[e rt ] = e rt (a 0 r n + a 1 r n a n 1 r + a n ) = 0 (7) a 0 r n + a 1 r n a n 1 r + a n = 0 Characteristic Equation (8) Since a 0 0, the characteristic equation has n zeros, say r 1, r 2,..., r n some of which may be equal. a 0 (r r 1 )(r r 2 )... (r r n ) = 0 (9) MATH 351 (Differential Equations) Ch 4 April 7, / 11

11 Real and Unequal Roots If the roots of the characteristic equation are real and no two are equal, then we have n distinct solutions e r 1t, e r 2t,, e rnt If these functions are linearly independent, then the general solution is y = c 1 e r 1t + c 2 e r 2t + + c n e rnt MATH 351 (Differential Equations) Ch 4 April 7, / 11

12 Complex Roots Since the coefficients a 0, a 1, a 2,..., a n are real numbers, if the characteristic equation has complex roots, they must occur in conjugate pairs, λ ± iµ. MATH 351 (Differential Equations) Ch 4 April 7, / 11

13 Repeated Roots Recalling the 2nd order linear equation with constant coefficients ay + by + cy = 0 y 1 = e r 1t and y 2 = te r 1t For an equation of order n: a 0 r n + a 1 r n a n 1 r + a n = 0 say r = r 1 real, has multiplicity s (where s n), then e r 1t, te r 1t, t 2 e r 1t,..., t s 1 e r 1t, are the corresponding solutions of the equation, and the general solution should include the linear combination c 1 e r 1t + c 2 te r 1t + c 3 t 2 e r 1t + + c s t s 1 e r 1t, MATH 351 (Differential Equations) Ch 4 April 7, / 11

14 Repeated Roots cont d complex root λ + iµ repeated s times, then λ iµ also repeated s, then e λt cos µt, te λt cos µt, t 2 e λt cos µt, t s 1 e λt cos µt, e λt sin µt, te λt sin µt, t 2 e λt sin µt,, t s 1 e λt sin µt, are the solutions of the equation and the general solution should include the linear combination c 1e λt cos µt + c 2e λt sin µt + c 3te λt cos µt + c 4te λt sin µt +c 5t 2 e λt cos µt + c 6t 2 e λt sin µt + + c 2s 1t s 1 e λt cos µt + c 2st s 1 e λt sin µt MATH 351 (Differential Equations) Ch 4 April 7, / 11

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