6 - Theory of Higher-Order Linear Differential Equations

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1 6 - Theory of Higher-Order Linear Differential Equations 6.1 Basic Theory of Linear Differential Equations Homework: p #1, 7, 15, 19 ü Introduction A linear differential equation of order n is an equation that can be written in the form (1) a n HxL y HnL HxL + a n-1 HxL y Hn-1L HxL + + a 0 HxL yhxl = bhxl where a 0 HxL, a 1 HxL,..., a n HxL and bhxl depend on x only. If we divide both sides by a n HxL, we obtain the standard form (2) y HnL HxL + p 1 HxL y Hn-1L HxL + + p n HxL yhxl = ghxl Theorem 1 - Existence and Uniqueness Suppose p 1 HxL,..., p n HxL and ghxl are each continuous on an interval Ha, bl that contains the point x 0. Then, for any choice of the initial values g 0, g 1,..., g n-1, there exists a unique solution yhxl on the whole interval Ha, bl to the initial value problem (3) y HnL HxL + p 1 HxL y Hn-1L HxL + + p n HxL yhxl = ghxl (4) yhx 0 L =g 0, y' Hx 0 L =g 1,..., y Hn-1L Hx 0 L =g n-1 We will often rewrite our differential equation in terms of the differential operator D. Recall that the D operator is another way of writing. So we can rewrite a differential equation as follows: x y'' - 3 y' - 4 y = 0ï D 2 y - 3 Dy-4 y = 0ïID 2-3 D - 4M@yD = 0 So we can define a linear operator L as (7) L@yD = y HnL HxL + p 1 HxL y Hn-1L HxL + + p n HxL yhxl = ID n + p 1 D n p n M@yD Then we can express the differential equation (2) as (8) L@yD = ghxl Because L is a linear operator, it satisfies the following: (9) L@y 1 + y y n D = L@y 1 D + L@y 2 D + + L@y n D and (10) L@c yd = cl@yd As a result of this linearity, if y 1,..., y m are solutions to the homogeneous equation L@yD = 0, then any linear combination C 1 y 1 + C 2 y C m y m is also a solution since (9) L@C 1 y 1 + C 2 y C m y m D = C 1 L@y 1 D + C 2 L@y 2 D + + C m L@y m D = 0

2 2 CH_06_notes.nb Definition 1 - Wronskian Let f 1,..., f n be any n functions that are Hn - 1L times differentiable. The function f 1 HxL f 2 HxL f n HxL f 1 HxL f 2 HxL f n HxL (16) W@ f 1,..., f n D := ª ª ª f Hn-1L 1 HxL f Hn-1L 2 HxL f Hn-1L n HxL is called the Wronskian of f 1,..., f n. Definition 2 - Linear Dependence of Functions The m functions f 1,..., f m are said to be linearly dependent on an interval I if at least one of them can be expressed as a linear combination of the others on I; equivalently, they are linearly dependent if there exist constants c 1, c 2,..., c m, not all zero such that (23) c 1 f 1 HxL + c 2 f 2 HxL + + c n f n HxL = 0 for all x in I. Otherwise, they are said to be linearly independent on I. Theorem 2 - Representation of Solutions (Homogeneous Case) Let y 1,..., y n be n solutions of Ha, bl of (17) y HnL HxL + p 1 HxL y Hn-1L HxL + + p n HxL yhxl = 0 where p 1,..., p n are continuous on Ha, bl. If at some point x 0 in Ha, bl these solutions satisfy (18) W@ f 1,..., f n DHx 0 L 0 then every solution of (17) on Ha, bl can be expressed in the form (19) yhxl = C 1 y 1 HxL + C 2 y 2 HxL + + C n y n HxL where C 1,..., C n are constants. Theorem 3 - Linear Dependence and the Wronskian If y 1,..., y n are n solutions to y HnL + p 1 y Hn-1L + + p n y = 0 on the interval Ha, bl with p 1,..., p n continuous on Ha, bl, then the following statements are equivalent: (i) y 1,..., y n are linearly dependent on Ha, bl. (ii) The Wronskian W@y 1,..., y n DHx 0 L is zero at some point x 0 in Ha, bl. (iii) The Wronskian W@y 1,..., y n DHxL is identically zero on Ha, bl. The contrapositives of these statements are also equivalent. (iv) y 1,..., y n are linearly independent on Ha, bl. (v) The Wronskian W@y 1,..., y n DHx 0 L is nonzero at some point x 0 in Ha, bl. (vi) The Wronskian W@y 1,..., y n DHxL is never zero on Ha, bl. Whenever (iv), (v), or (vi) is met, 8y 1,..., y n < is called a fundamental solution set for (17) on Ha, bl. Theorem 4 - Representation of Solutions (Nonhomogeneous Case) Let y p HxL be a particular solution of the nonhomogeneous equation (27) y HnL HxL + p 1 HxL y Hn-1L HxL + + p n HxL yhxl = ghxl on the interval Ha, bl with p 1,..., p n are continuous on Ha, bl and let 8y 1,..., y n < be a fundamental solution set for the corresponding homogeneous equation (28) y HnL HxL + p 1 HxL y Hn-1L HxL + + p n HxL yhxl = 0. Then every solution of (27) on the interval Ha, bl can be expressed in the form (29) yhxl = y p HxL + C 1 y 1 HxL + C 2 y 2 HxL + + C n y n HxL.

3 CH_06_notes.nb 3 ü Example Use the Wronskian to verify that the functions 8 x, sin 2 x, cos 2 x< form a fundamental solution set for the differential equation and find a general solution. 3 y - 2 y + 4 y - 4 y = 0. x 3 x 2 x

4 4 CH_06_notes.nb 6.2 Homogeneous Linear Equations with Constant Coefficients Homework: p #1, 5, 9, 13, 15, 19 ü Introduction As before, we will first consider the homogeneous nth-order linear differential equation (1) a n y HnL HxL + a n-1 y Hn-1L HxL + + a 0 yhxl = 0 where a n 0, a n-1,..., a 0 are real constants. Then (1) has solutions defined for all x in H-, L. We must find n linearly independent solutions y 1,..., y n, then our general solution is given by (2) yhxl = C 1 y 1 HxL + C 2 y 2 HxL + + C n y n HxL where C 1,..., C n are constants. As before, we will assume y = rx is a solution, and substituting it and its derivatives into the differential equation (1) we obtain the auxiliary (or characteristic) equation (6) PHrL = a n r n + a n-1 r n a 0 = 0 and as before we must find the roots of this equation. Again, three cases to consider. ü Distinct Real Roots If the roots of the auxiliary equation (6) are the distinct real numbers r 1, r 2,..., r n, then n solutions to equation (1) are given by y 1 HxL = r 1 x, y 2 HxL = r 2 x,..., y n HxL = r n x and the general solution to (1) is yhxl = C 1 r 1 x + C 2 r 2 x + + C n r n x where C 1,..., C n are constants. ü Example Find a general solution to y''' + 3 y'' - 4 y' - 12 y = 0. ü Complex Roots If a+bâ (a, b real) is a complex root of the auxiliary equation (6), then the complex conjugate a-bâ is also a root. So two linearly independent solutions to (1) are a x cos b x and a x sin b x. ü Example Find a general solution to y''' - y'' + 2 y = 0.

5 CH_06_notes.nb 5 ü Repeated Roots If r 1 is a root of (6) of multiplicity m, then m linearly solutions are r 1 x, x r 1 x,..., x m-1 r 1 x. ü Example Find a general solution to y''' - 6 y'' + 12 y' - 8 y = 0. ü Example Find a general solution to y H4L + 8 y'' + 16 y = 0.

6 6 CH_06_notes.nb 6.3 Undetermined Coefficients and the Annihilator Method Homework: p #1, 5, 9, 31 ü Introduction Now we consider solving a nonhomogeneous equation a n y HnL HxL + a n-1 y Hn-1L HxL + + a 0 yhxl = ghxl As before, first find the solutions to the corresponding homogeneous equation a n y HnL HxL + a n-1 y Hn-1L HxL + + a 0 yhxl = 0 then find a particular solution to the nonhomogeneous equation a n y HnL HxL + a n-1 y Hn-1L HxL + + a 0 yhxl = ghxl To do this, we will rely on the techniques discussed in Sections 4.4 and 4.5. ü Example Find a general solution to y''' + y'' = 3 x + 4 x 2.