A Second Course in Elementary Differential Equations

Transcription

1 A Second Course in Elementary Differential Equations Marcel B Finan Arkansas Tech University c All Rights Reserved August 3, 23

2 Contents 28 Calculus of Matrix-Valued Functions of a Real Variable 4 29 nth Order Linear Differential Equations:Existence and Uniqueness 9 3 The General Solution of nth Order Linear Homogeneous Equations 26 3 Fundamental Sets and Linear Independence Higher Order Homogeneous Linear Equations with Constant Coefficients Non Homogeneous nth Order Linear Differential Equations 5 34 Existence and Uniqueness of Solution to Initial Value First Order Linear Systems 6 35 Homogeneous First Order Linear Systems First Order Linear Systems: Fundamental Sets and Linear Independence Homogeneous Systems with Constant Coefficients Homogeneous Systems with Constant Coefficients: Complex Eigenvalues Homogeneous Systems with Constant Coefficients: Repeated Eigenvalues 2 4 Nonhomogeneous First Order Linear Systems 4 4 Solving First Order Linear Systems with Diagonalizable Constant Coefficients Matrix Solving First Order Linear Systems Using Exponential Matrix 33 2

3 43 The Laplace Transform: Basic Definitions and Results Further Studies of Laplace Transform 5 45 The Laplace Transform and the Method of Partial Fractions64 46 Laplace Transforms of Periodic Functions 7 47 Convolution Integrals 8 48 The Dirac Delta Function and Impulse Response Solving Systems of Differential Equations Using Laplace Transform 97 5 Numerical Methods for Solving First Order Linear Systems: Euler s Method 24 3

4 28 Calculus of Matrix-Valued Functions of a Real Variable In establishing the existence result for second and higher order linear differential equations one transforms the equation into a linear system and tries to solve such a system This procedure requires the use of concepts such as the derivative of a matrix whose entries are functions of t, the integral of a matrix, and the exponential matrix function Thus, techniques from matrix theory play an important role in dealing with systems of differential equations The present section introduces the necessary background in the calculus of matrix functions Matrix-Valued Functions of a Real Variable A matrix A of dimension m n is a rectangular array of the form a a 2 a n A = a 2 a 22 a 2n a m a m2 a mn where the a ij s are the entries of the matrix, m is the number of rows, n is the number of columns The zero matrix is the matrix whose entries are all The n n identity matrix I n is a square matrix whose main diagonal consists of s and the off diagonal entries are all A matrix A can be represented with the following compact notation A = (a ij ) The entry a ij is located in the ith row and jth column Example 28 Consider the matrix Find a 22, a 32, and a 23 A(t) = Solution The entry a 22 is in the second row and second column so that a 22 = 2 Similarly, a 32 = 2 and a 23 = 4

5 An m n array whose entries are functions of a real variable defined on a common interval is called a matrix function Thus, the matrix a (t) a 2 (t) a 3 (t) A(t) = a 2 (t) a 22 (t) a 23 (t) a 3 (t) a 32 (t) a 33 (t) is a 3 3 matrix function whereas the matrix x (t) x(t) = x 2 (t) x 3 (t) is a 3 matrix function also known as a vector-valued function We will denote an m n matrix function by A(t) = (a ij (t)) where a ij (t) is the entry in the ith row and jth coloumn Arithmetic of Matrix Functions All the familiar rules of matrix arithmetic hold for matrix functions as well (i) Equality: Two m n matrices A(t) = (a ij (t)) and B(t) = (b ij (t)) are said to be equal if and only if a ij (t) = b ij (t) for all i m and j n That is, two matrices are equal if and only if all corresponding elements are equal Notice that the matrices must be of the same dimension Example 282 Solve the following matrix equation for a, b, c, and d [ ( ) a b b + c 8 = 3d + c 2a 4d 7 6 Solution Equating corresponding entries we get the system a b = 8 b + c = c + 3d = 7 2a 4d = 6 Adding the first two equations to obtain a + c = 9 Adding 4 times the third equation to 3 times the last equation to obtain 6a + 4c = 46 or 3a + 2c = 23 5

6 Solving the two equations in a and c one finds a = 5 and c = 4 Hence, b = 3 and d = (ii) Addition: If A(t) = (a ij (t)) and B(t) = (b ij (t) are m n matrices then the sum is a new m n matrix obtained by adding corresponding elements (A + B)(t) = A(t) + B(t) = (a ij (t) + b ij (t)) Matrices of different dimensions cannot be added (iii) Subtraction: Let A(t) = (a ij (t)) and B(t) = (b ij (t)) be two m n matrices Then the difference (A B)(t) is the new matrix obtained by subtracting corresponding elements,that is (A B)(t) = A(t) B(t) = (a ij (t) b ij (t)) (iv) Scalar Multiplication: If α is a real number and A(t) = (a ij (t)) is an m n matrix then (αa)(t) is the m n matrix obtained by multiplying the entries of A by the number α; that is, (αa)(t) = αa(t) = (αa ij (t)) (v) Matrix Multiplication: If A(t) is an m n matrix and B(t) is an n p matrix then the matrix AB(t) is the m p matrix where AB(t) = (c ij (t)) c ij (t) = n a ik (t)b kj (t) k= That is the c ij entry is obtained by multiplying componentwise the ith row of A(t) by the jth column of B(t) It is important to realize that the order of the multiplicands is significant, in other words AB(t) is not necessarily equal to BA(t) In mathematical terminology matrix multiplication is not commutative Example 283 A = [ [ 2, B = 3 4 Show that AB BA Hence, matrix multiplication is not commutative 6

7 Solution Using the definition of matrix multiplication we find [ [ AB =, BA = Hence, AB BA (vi) Inverse: An n n matrix A(t) is said to be invertible if and only if there is an n n matrix B(t) such that AB(t) = BA(t) = I where I is the matrix whose main diagonal consists of the number and elsewhere We denote the inverse of A(t) by A (t) Example 284 Find the inverse of the matrix [ a b A = c d given that ad bc The quantity ad bc is called the determinant of A and is denoted by deta Solution Suppose that Then [ x y z t This implies that Hence, [ x y A = z t [ a b = c d [ ax + cy bx + dy = az + ct bz + dt ax + cy = bx + dy = az + ct = bz + dt = [ [ Applying the method of elimination to the first two equations we find 7

8 x = d and y = ad bc b ad bc Applying the method of elimination to the last two equations we find Hence, z = c and t = ad bc a ad bc [ A d b = ad bc c a Norm of a Vector Function The norm of a vector function will be needed in the coming sections In one dimension a norm is known as the absolute value In multidimenesion, we define the norm of a vector function x with components x, x 2,, x n by x = x + x x n From this definition one notices the following properties: (i) If x = then x + x x n = and this implies that x = x 2 = = x n = Hence, x = (ii) If α is a scalar then αx = αx + αx αx n = α ( x + x x n ) = α x (iii) If x is vector function with components x, x 2,, x n and y with components y, y 2,, y n then x + y = x + y + x 2 + y x n + y n ( x + x x n ) + ( y + y y n ) = x + y Limits of Matrix Functions If A(t) = (a ij (t)) is an m n matrix such that lim t t a ij (t) = L ij exists for all i m and j n then we define lim A(t) = (L ij ) t t Example 285 Suppose that Find lim t A(t) [ t A(t) = 2 5t t 3 2t 3 8

9 Solution lim A(t) = t [ limt (t 2 5t) lim t t 3 lim t 2t lim t 3 = [ If one or more of the component function limits does not exist, then the limit of the matrix does not exist For example, if [ t t A(t) = e t then lim t A does not exist since lim t does not exist t We say that A(t) is continuous at t = t if lim A(t) = A(t ) t t Example 286 Show that the matrix is continuous at t = [ t t A(t) = e t Solution Since [ 2 /2 lim A(t) = t e 2 = A() we have A(t) is continuous at t = Most properties of limits for functions of a single variable are also valid for limits of matrix functions Matrix Differentiation Let A(t) be an m n matrix such that each entry is a differentiable function of t We define the derivative of A(t) to be provided that the limit exists A (t) = lim h A(t + h) A(t) h 9

10 Example 287 Let [ a (t) a A(t) = 2 (t) a 2 (t) a 22 (t) where the entries a, a 2, a 2, and a 22 are differentiable Find A (t) Solution We have A A(t+h) A(t) (t) = lim h h [ = = lim h a (t+h) a (t) h lim h a 2 (t+h) a 2 (t) h [ a (t) a 2(t) a 2(t) a 22(t) lim h a 2 (t+h) a 2 (t) h lim h a 22 (t+h) a 22 (t) h It follows from the previous example that the derivative of a matrix function is the matrix of derivatives of its component functions From this fact one can check easily the following two properties of differentiation: (i) If A(t) and B(t) are two m n matrices with both of them differentiable then the matrix (A + B)(t) is also differentiable and (A + B) (t) = A (t) + B (t) (ii) If A(t) is an m n differentiable matrix and B(t) is an n p differentiable matrix then the product matrix AB(t) is also differentiable and Example 288 Write the system in matrix form (AB) (t) = A (t)b(t) + A(t)B (t) y = a (t)y (t) + a 2 (t)y 2 (t) + a 3 (t)y 3 (t) + g (t) y 2 = a 2 (t)y (t) + a 22 (t)y 2 (t) + a 23 (t)y 3 (t) + g 2 (t) y 3 = a (t)y (t) + a 2 (t)y 2 (t) + a 3 (t)y 3 (t) + g 3 (t)

11 Solution Let y(t) = y (t) y 2 (t) y 3 (t), A(t) = a (t) a 2 (t) a 3 a 2 (t) a 22 (t) a 23 a 3 a 32 a 33 Then the given system can be written in the matrix form y (t) = A(t)y(t) + g(t), g(t) = g (t) g 2 (t) g 3 (t) Matrix Integration: Since the derivative of a matrix function is a matrix of derivatives, it should not be surprising that antiderivatives of a matrix function can be evaluated by performing the corresponding antidifferentiation operations upon each component of the matrix function That is, if A(t) is the m n matrix A(t) = a (t) a 2 (t) a n (t) a 2 (t) a 22 (t) a 2n (t) a m (t) a m2 (t) a mn (t) then A(t)dt = a (t)dt a2 (t)dt am (t)dt a2 (t)dt a22 (t)dt am2 (t)dt an (t)dt a2n (t)dt amn (t)dt Example 289 Determine the matrix function A(t) if [ 2t A (t) = cos t 3t 2 Solution We have [ t A(t) = 2 + c t + c 2 sin t + c 2 t 3 + c 22 = [ t 2 t sin t t 3 [ c c + 2 c 2 c 22

12 Finally, we conclude this section by showing that t x(s)ds t t x(s) ds t To see this, t t x(s)ds = t t t x (s)ds t x 2 (s)ds t t x n (s)ds = t t x (s)ds + t t x 2 (s)ds + t t x n (s)ds t t x (s) ds + t t x 2 (s) ds + + t t x n (s) ds = t t ( x + x x n )ds = t t x(s) ds 2

13 Practice Problems Problem 28 Consider the following matrices [ t t 2 A(t) = 2 2t + (a) Find 2A(t) - 3tB(t) (b) Find A(t)B(t) - B(t)A(t) (c) Find A(t)c(t) (d) Find det(b(t)a(t)), B(t) = [ t t + 2 [ t +, c(t) = Problem 282 Determine all values t such that A(t) is invertible and, for those t-values, find A (t) [ t + t A(t) = t t + Problem 283 Determine all values t such that A(t) is invertible and, for those t-values, find A (t) [ sin t cos t A(t) = sin t cos t Problem 284 Find Problem 285 Find [ sin t 3 t cos t lim t t+ t e 3t 2t sec t t 2 [ te t tan t lim t t 2 2 e sin t Problem 286 Find A (t) and A (t) if A(t) = [ sin t 3t t

14 Problem 287 Express the system in the matrix form y = t 2 y + 3y 2 + sec t y 2 = (sin t)y + ty 2 5 y (t) = A(t)y(t) + g(t) Problem 288 Determine A(t) where A (t) = Problem 289 Determine A(t) where [ t A (t) = [ 2t cos t 3t 2 [, A() = 2 [ 2 5, A() = 2 [ 2, A () = 2 3 Problem 28 Calculate A(t) = t B(s)ds where [ e B(s) = s cos 2πs 6s sin 2πs Problem 28 Construct a 2 2 a nonconstant matrix function A(t) such that A 2 (t) is a constant matrix Problem 282 (a) Construct a 2 2 differentiable matrix function A(t) such that d dt A2 (t) 2A d dt A(t) That is, the power rule is not true for matrix functions (b) What is the correct formula relating A 2 (t) to A(t) and A (t)? 4

15 Problem 283 Transform the following third-order equation into a first order system of the form y 3ty + (sin 2t)y = 7e t x (t) = Ax(t) + b(t) Problem 284 By introducing new variables x and x 2, write y 2y + = t as a system of two first order linear equations of the form x + Ax = b Problem 285 Write the differential equation y + 4y + 4y = as a first order system Problem 286 Write the differential equation y + ky + (t )y = as a first order system Problem 287 Change the following second-order equations to a first-order system y 5y + ty = 3t 2, y() =, y () = Problem 288 Consider the following system of first-order linear equations x = 3 2 x Find the second-order linear differential equation that x satisfies The Determinant of a Matrix The determinant of a matrix function is the same as the determinant with constant entries So we will introduce the definition of determinant of a matrix with constant entries A permutation of the set S = {, 2,, n} is an arrangement of the elements of S in some order without omissions or repetitions We write σ = (σ()σ(2) σ(n)) In terms of functions, a permutation is a one-toone function from S onto S 5

16 Let S n denote the set of all permutations on S How many permutations are there in S n? We have n positions to be filled by n numbers For the first position, there are n possibilities For the second there are n possibilities, etc Thus, according to the multiplication rule of counting there are n(n )(n 2) 2 = n! permutations Is there a way to list all the permutations of S n? The answer is yes and one can find the permutations by using a permutation tree which we describe in the following example Problem 289 List all the permutations of S = {, 2, 3, 4} An inversion is said to occur whenever a larger integer precedes a smaller one If the number of inversions is even (resp odd) then the permutation is said to be even (resp odd) We define the sign of a permutation to be a function sgn with domain S n and range {, } such that sgn(σ) = if σ is odd and sgn(σ) = + if σ is even For example, the permutation in S 6 defined by σ() = 3, σ(2) = 6, σ(3) = 4, σ(5) = 2, σ(6) = is an even permuatation since the inversions are (6,),(6,3),(6,4),(6,5),(6,2),(3,2),(4,2), and (5,2) Let A be an n n matrix An elementary product from A is a product of n entries from A, no two of which come from the same row or same column Problem 282 List all elementary products from the matrices (a) ( ) a a 2, a 2 a 22 (b) a a 2 a 3 a 2 a 22 a 23 a 3 a 32 a 33 6

17 Let A be an n n matrix Consider an elementary product of entries of A For the first factor, there are n possibilities for an entry from the first row Once selected, there are n possibilities for an entry from the second row for the second factor Continuing, we find that there are n! elementary products They are the products of the form a σ() a 2σ(2) a nσ(n), where σ is a permutation of {, 2,, n}, ie a member of S n Let A be an n n matrix Then we define the determinant of A to be the number det(a) = sgn(σ)a σ() a 2σ(2) a nσ(n) where the sum is over all permutations σ of {, 2,, n} Problem 282 Find det(a) if (a) ( a a A = 2 a 2 a 22 ), (b) A = a a 2 a 3 a 2 a 22 a 23 a 3 a 32 a 33 The following theorem is of practical use It provides a technique for evaluating determinants by greatly reducing the labor involved Theorem 28 Let A be an n n matrix (a) Let B be the matrix obtained from A by multiplying a row or a column by a scalar c Then det(b) = cdeta (b) Let B be the matrix obtained from A by interchanging two rows or two columns of A Then det(b) = det(a) (c) If A has two identical rows or columns then its determinant is zero (d) Let B be the matrix obtained from A by adding c times a row (or a column) to another row (column) Then det(b) = det(a) (e) The determinant of the product of two n n matrices is the product of their determinant (g) If B is the matrix whose columns are the rows of A then det(b) = det(a) 7

18 The proof of this theorem can be found in any textbook in elementary linear algebra 8

19 29 nth Order Linear Differential Equations:Existence and Uniqueness In the following three sections we carry the basic theory of second order linear differential equations to nth order linear differential equation y (n) + p n (t)y (n ) + + p (t)y + p (t)y = g(t) where the functions p, p,, p n and g(t) are continuous functions for a < t < b If g(t) is not identically zero, then this equation is said to be nonhomogeneous; if g(t) is identically zero, then this equation is called homogeneous Existence and Uniqueness of Solutions We begin by discussing the existence of a unique solution to the initial value problem y (n) + p n (t)y (n ) + + p (t)y + p (t)y = g(t) y(t ) = y, y (t ) = y,, y (n ) (t ) = y (n ), a < t < b The following theorem is a generalization to Theorems 32 and 5 Theorem 29 The nonhomogeneous nth order linear differential equation with initial conditions y (n) + p n (t)y (n ) + + p (t)y + p (t)y = g(t) () y(t ) = y, y (t ) = y,, y (n ) (t ) = y (n ), a < t < b (2) has a unique solution in a < t < b Proof Existence: The existence of a local solution is obtained here by transforming the problem into a first order system This is done by introducing the variables x = y, x 2 = y,, x n = y (n ) 9

20 In this case, we have x = x 2 x 2 = x 3 = x n = x n x n = p n (t)x n + p (t)x 2 p (t)x + g(t) Thus, we can write the problem as a system: x x 2 x 3 + x n or in compact form where p p p 2 p 3 p n x x 2 x 3 x n = g(t) x (t) = A(t)x(t) + b(t), x(t ) = y (3) A(t) = x(t) = x x 2 x 3 x n p p p 2 p 3 p n y y, b(t) =, y = y (n ) g(t) Note that if y(t) is a solution of () then the vector-valued function y y x(t) = y (n ) 2

21 is a solution to (3) Conversely, if the vector x x 2 x(t) = x 3 x n is a solution of (3) then x = x 2, x = x 3,, x (n ) = x n Hence, x (n) = x n = p n (t)x n p n 2 (t)x n p (t)x + g(t) or x (n) + p n (t)x (n ) + p n 2 (t)x (n 2) + + p (t)x = g(t) which means that x is a solution to () Next, we start by reformulating (3) as an equivalent integral equation Integration of both sides of (3) yields t t x (s)ds = t t [A(s)x(s) + b(s)ds (4) Applying the Fundamental Theorem of Calculus to the left side of (4) yields x(t) = x(t ) + t t [A(s)x(s) + b(s)ds (5) Thus, a solution of (5) is also a solution to (3) and vice versa To prove the existence and uniqueness, we shall use again the method of successive approximation as described in Theorem 8 Letting y y y = y (n ) we can introduce Picard s iterations defined recursively as follows: y (t) y y (t) = y + t t [A(s)y (s) + b(s)ds y 2 (t) = y + t t [A(s)y (s) + b(s)ds y N (t) = y + t t [A(s)y N (s) + b(s)ds 2

22 Let y N (t) = y,n y 2,N y n,n For i =, 2,, n, we are going to show that the sequence {y i,n (t)} converges uniformly to a function y i (t) such that y(t) (with components y, y 2,, y n )is a solution to (5) and hence a solution to (3) Let [c, d be a closed interval containing t and contained in (a, b) Then there exist constants k, k,, k n such that max p (t) k, c t d This implies that max p (t) k,, max p n (t) k n c t d c t d A(t)x(t) = x 2 + x x n + p x + p x p n x n k x + ( + k ) x ( + k n 2 ) x n + k n x n K x for all c t d, where we define and where It follows that for i n But where y = y + y y n K = k + ( + k ) + + ( + k n 2 ) + k n y i,n y i,n y N y N t t A(s) (y N y N 2 ) ds K t t y N y N 2 ds An easy induction yields that y y t t A(s) y + b(s) ds M(t t ) M = K y + max c t d b(t) y N+ y N MK N (t t ) N+ N! MK N (b a)n+ N! 22

23 Since N= N (b a)n+ MK N! = M(b a)(e K(b a) ) by Weierstrass M-test we conclude that the series N= [y i,n y i,n converges uniformly for all c t d But N y i,n (t) = [y i,k+ (t) y i,k (t) + y i, k= Thus, the sequence {y i,n } converges uniformly to a function y i (t) for all c t d The function y i (t) is a continuous function (a uniform limit of a sequence of continuous function is continuous) Also we can interchange the order of taking limits and integration for such sequences Therefore y(t) = lim N y N (t) t = y + lim N t (A(s)y N + b(s))ds = y + t t lim N (A(s)y N + b(s))ds = y + t t (A(s)y + b(s))ds This shows that y(t) is a solution to the integral equation (5) and therefore a solution to (3) Uniqueness: The uniqueness follows from Gronwall Inequality (See Problem 8) Suppose that y(t) and z(t) are two solutions to the initial value problem, it follows that for all a < t < b we have y(t) z(t) Letting u(t) = y(t) z(t) we have u(t) t t K y(s) z(s) ds t t Ku(s)ds so that by Gronwall s inequality u(t) and therefore y(t) = z(t) for all a < t < b This completes a proof of the theorem 23

24 Example 29 Find the largest interval where (t 2 6)y (4) + 2y + t 2 y = sec t, y(3) =, y (3) = 3, y (3) = is guaranteed to have a unique solution Solution We first put it into standard form y (4) + 2 t 2 6 y + t2 t 2 6 y = sect t 2 6 The coefficient functions are continuous for all t ±4 and t (2n + ) π 2 Since t = 3, the largest interval where the given initial value problem is guaranteed to have a unique solution is the ineterval π 2 < t < 4 24

25 Practice Problems For Problems , use Theorem 29 to find the largest interval a < t < b in which a unique solution is guaranteed to exist Problem 29 y t 2 9 y + ln (t + )y + (cos t)y =, y() =, y () = 3, y () = Problem 292 Problem 293 y + t + y + (tan t)y =, y() =, y () =, y () = 2 y t y + ln (t 2 + )y + (cos t)y =, y() =, y () = 3, y () = Problem 294 Determine the value(s) of r so that y(t) = e rt is a solution to the differential equation y 2y y + 2y = Problem 295 Transform the following third-order equation into a first order system of the form y 3ty + (sin 2t)y = 7e t x (t) = Ax(t) + b(t) 25

26 3 The General Solution of nth Order Linear Homogeneous Equations In this section we consider the question of solving the homogeneous equation y (n) + p n (t)y (n ) + + p (t)y + p (t)y = (6) where p (t), p (t),, p n (t) are continuous functions in the interval a < t < b The next theorem shows that any linear combination of solutions to the homogeneous equation is also a solution In what follows and for the simplicity of arguments we introduce the function L defined by L[y = y (n) + p n (t)y (n ) + + p (t)y + p (t)y Theorem 3 (Linearity) If y and y 2 are n times differentiable and α and α 2 are scalars then L satisfies the property Proof Indeed, we have L[α y + α 2 y 2 = α L[y + α 2 L[y 2 L[α y + α 2 y 2 = (α y + α 2 y 2 ) (n) + p n (t)(α y + α 2 y 2 ) (n ) + + p (t)(α y + α 2 y 2 ) = (α y (n) + α p n (t)y (n ) + + α p (t)y + α p (t)y ) + (α 2 y (n) 2 + α 2 p n (t)y (n ) α 2 p (t)y 2 + α 2 p (t)y 2 ) = α (y (n) + p n (t)y (n ) + + p (t)y + p (t)y ) + α 2 (y (n) 2 + p n (t)y (n ) p (t)y 2 + p (t)y 2 ) = α L[y + α 2 L[y 2 The above property applies for any number of functions An important consequence of this theorem is the following result Corollary 3 (Principle of Superposition) If y, y 2,, y r satisfy the homogeneous equation (6) and if α, α 2,, α r are any numbers, then y(t) = α y + α 2 y α r y r also satisfies the homogeneous equation (6) 26

27 Proof Since y, y 2,, y r are solutions to (6), L[y = L[y 2 = = L[y r = Now, using the linearity property of L we have L[α y + α 2 y α r y r = α L[y + α 2 L[y α r L[y r = = The principle of superposition states that if y, y 2,, y r are solutions to (6) then any linear combination is also a solution The next question that we consider is the question of existence of n solutions y, y 2,, y n of equation (6) such that every solution to (6) can be written as a linear combination of these functions We call such a set a functions a fundamental set of solutions Note that the number of solutions comprising a fundamental set is equal to the order of the differential equation Also, note that the general solution to (6) is then given by y(t) = c y (t) + c 2 y 2 (t) + + c n y n (t) Next, we consider a criterion for testing n solutions for a fundamental set For that we first introduce the following definition: For n functions y, y 2,, y n, we define the Wronskian of these functions to be the determinant W (t) = y (t) y 2 (t) y n (t) y (t) y 2(t) y n(t) y (t) y 2(t) y n(t) y (n ) (t) y (n ) 2 (t) y n (n ) (t) Theorem 32 (Criterion for identifying fundamental sets) Let y (t), y 2 (t),, y n (t) be n solutions to the homogeneous equation (6) If there is a a < t < b such that W (t ) then {y, y 2,, y n } is a fundamental set of solutions Proof We need to show that if y(t) is a solution to (6) then we can write y(t) as a linear combination of y, y 2 (t),, y n (t) That is y(t) = c y (t) + c 2 y 2 (t) + + c n y n 27

28 So the problem reduces to finding the constants c, c 2,, c n These are found by solving the following linear system of n equations in the unknowns c, c 2,, c n : c y (t ) + c 2 y 2 (t ) + + c n y n (t ) = y(t ) c y (t ) + c 2 y 2(t ) + + c n y n(t ) = y (t ) = c y (n ) (t) + c 2 y (n ) 2 (t) + + c n y n (n ) (t) = y (n ) (t ) Solving this system using Cramer s rule we find c i = W i(t ) W (t ), i n where W i (t ) = y (t ) y 2 (t ) y(t ) y n (t ) y (t ) y 2(t ) y (t ) y n(t ) y (t ) y 2(t ) y (t ) y n(t ) y (n ) (t ) y (n ) 2 (t ) y (n ) (t ) y n (n ) (t ) That is, W i (t ) is the determinant of W with the ith column being replaced by the right-hand column of the above system Note that c, c 2,, c n exist since W (t ) As a first application to this result, we establish the existence of fundamental sets Theorem 33 The linear homogeneous differential equation y (n) + p n (t)y (n ) + + p (t)y + p (t)y = where p n (t),, p (t), p (t) are continuous functions in a < t < b has a fundamental set {y, y 2,, y n } Proof Pick a t in the interval a < t < b and consider the n initial value problems y (n) +p n (t)y (n ) + +p (t)y +p (t)y =, y(t ) =, y (t ) =, y (t ) =,, y (n ) (t ) = 28

29 y (n) +p n (t)y (n ) + +p (t)y +p (t)y =, y(t ) =, y (t ) =, y (t ) =,, y (n ) (t ) = y (n) +p n (t)y (n ) + +p (t)y +p (t)y =, y(t ) =, y (t ) =, y (t ) =,, y (n ) (t ) = y (n) +p n (t)y (n ) + +p (t)y +p (t)y =, y(t ) =, y (t ) =, y (t ) =,, y (n ) (t ) = Then by Theorem 27, we can find unique solutions {y, y 2,, y n } This set is a fundamental set by the previous theorem since W (t) = = Theorem 32 says that if one can find a < t < b such that W (t ) then the set {y, y 2,, y n } is a fundamental set of solutions The following theorem shows that the condition W (t ) implies that W (t) for all t in the interval (a, b) That is, the theorem tells us that we can choose our test point t on the basis of convenience-any test point t will do Theorem 34 (Abel s) Let y (t), y 2 (t),, y n be n solutions to equation (6) Then () W (t) satisfies the differential equation W (t) + p n (t)w (t) = ; (2) If t is any point in (a, b) then W (t) = W (t )e t t p n (s)ds Thus, if W (t ) then W (t) for all a < t < b Proof () By introducing the variables x = y, x 2 = y, x 3 = y,, x n = y (n ) we can write the differential equation as a first order system in the form where A(t) = x (t) = A(t)x(t) p p p 2 p 3 p n 29, x(t) = x x 2 x 3 x n

30 We will show in Section 33 that for any linear system of the form we have In our case so that x (t) = A(t)x(t) W (t) = (a + a a nn )W (t) a + a a nn = p n (t) W (t) + p n (t)w (t) = (2) The previous differential equation can be solved by the method of integrating factor to obtain W (t) = W (t )e t t p n (s)ds Example 3 Use the Abel s formula to find the Wronskian of the DE: ty + 2y t 3 y + e t2 y = Solution The original equation can be written as By Abel s formula the Wronskian is Example 32 Consider the linear system where y + 2 t y t 2 y + et2 t y = W (t) = Ce 2 t dt = C t 2 x (t) = A(t)x(t) [ a a A = 2 a 2 a 22 Show that for any two solutions Y and Y 2 we have W (t) = (a + a 22 )W (t) 3

31 Solution Suppose that Y = [ u u 2 [ v, Y 2 = v 2 are solutions to the given system Then we have W d (t) = dt u v u 2 v 2 = u v u 2 v 2 + u v u 2 v 2 But u v u 2 v 2 = a u + a 2 u 2 a v + a 2 v 2 u 2 v 2 = a W (t) and u v u 2 v 2 = u v a 2 u + a 22 u 2 a 2 v + a 22 v 2 = a 22W (t) It follows that W (t) = (a + a 22 )W (t) We end this section by showing that the converse of Theorem 32 is also true Theorem 35 If {y, y 2,, y n } is a fundamental set of solutions to y (n) + p n (t)y (n ) + + p (t)y + p (t)y = where p n (t),, p (t), p (t) are continuous functions in a < t < b then W (t) for all a < t < b Proof Let t be any point in (a, b) By Theorem 27, there is a unique solution y(t) to the initial value problem y (n) +p n (t)y (n ) + +p (t)y +p (t)y =, y(t ) =, y (t ) =,, y (n ) (t ) = 3

32 Since {y, y 2,, y n } is a fundamental set, there exist unique constants c, c 2,, c n such that c y (t) + c 2 y 2 (t) c n y n (t) = y(t) c y (t) + c 2 y 2(t) c n y n(t) = y (t) c y (n ) (t) + c 2 y (n ) 2 (t) c n y n (n ) (t) = y (n ) (t) for all a < t < b In particular for t = t we obtain the system c y (t) + c 2 y 2 (t) c n y n (t) = c y (t) + c 2 y 2(t) c n y n(t) = c y (n ) (t) + c 2 y (n ) 2 (t) c n y n (n ) (t) = This system has a unique solution (c, c 2,, c n ) where c i = W i W (t ) and W i is the determinant W with the ith column replaced by the column Note that for c, c 2,, c n to exist we must have W (t ) By Abel s theorem we conclude that W (t) for all a < t < b 32

33 Practice Problems In Problems 3-33, show that the given solutions form a fundamental set for the differential equation by computing the Wronskian Problem 3 Problem 32 y y =, y (t) =, y 2 (t) = e t, y 3 (t) = e t y (4) + y =, y (t) =, y 2 (t) = t, y 3 (t) = cos t, y 4 (t) = sin t Problem 33 t 2 y + ty y =, y (t) =, y 2 (t) = ln t, y 3 (t) = t 2 Use the fact that the solutions given in Problems 3-33 for a fundamental set of solutions to solve the following initial value problems Problem 34 Problem 35 y y =, y() = 3, y () = 3, y () = y (4) + y =, y( pi 2 ) = 2 + π, y ( π 2 ) = 3, y ( π 2 ) = 3, y ( π 2 ) = Problem 36 t 2 y + ty y =,, y() =, y () = 2, y () = 6 33

34 Problem 37 In each question below, show that the Wronskian determinant W (t) behaves as predicted by Abel s Theorem That is, for the given value of t, show that W (t) = W (t )e t t p n (s)ds (a) W (t) found in Problem 3 and t = (b) W (t) found in Problem 32 and t = (c) W (t) found in Problem 33 and t = 2 Problem 38 Determine W (t) for the differential equation y +(sin t)y +(cos t)y +2y = such that W () = Problem 39 Determine W (t) for the differential equation t 3 y 2y = such that W () = 3 Problem 3 Consider the initial value problem y y =, y() = α, y () = β, y () = 4 The general solution of the differential equation is y(t) = c + c 2 e t + c 3 e t (a) For what values of α and β will lim t y(t) =? (b) For what values α and β will the solution y(t) be bounded for t, ie, y(t) M for all t and for some M >? Will any values of α and β produce a solution y(t) that is bounded for all real number t? Problem 3 Consider the differential equation y + p 2 (t)y + p (t)y = on the interval < t < Suppose it is known that the coefficient functions p 2 (t) and p (t) are both continuous on < t < Is it possible that y(t) = c + c 2 t 2 + c 3 t 4 is the general solution for some functions p (t) and p 2 (t) continuous on < t <? (a) Answer this question by considering only the Wronskian of the functions, t 2, t 4 on the given interval (b) Explicitly determine functions p (t) and p 2 (t) such that y(t) = c +c 2 t 2 + c 3 t 4 is the general solution of the differential equation Use this information, in turn, to provide an alternative answer to the question 34

35 Problem 32 (a) Find the general solution to y = (b) Using the general solution in part (a), construct a fundamental set {y (t), y 2 (t), y 3 (t)} satisfying the following conditions y () =, y () =, y () = y 2 () =, y () =, y () = y () =, y () =, y () = 35

36 3 Fundamental Sets and Linear Independence In Section 3 we established the existence of fundamental sets There remain two questions that we would like to answer The first one is about the number of fundamental sets That is how many fundamental sets are there It turns out that there are more than one In this case, our second question is about how these sets are related In this section we turn our attention to these questions We start this section by the following observation Suppose that the Wronskian of n solutions {y, y 2,, y n } to the nth order linear homogeneous differential equation is zero In terms of linear algebra, this means that one of the columns of W can be written as a linear combination of the remaining columns For the sake of argument, suppose that the last column is a linear combination of the remaining columns: y n y n y (n ) n This implies that = c y y y (n ) + c 2 y 2 y 2 y (n ) c n y n (t) = c y (t) + c 2 y 2 (t) + + c n y n (t) y n y n y (n ) n Such a relationship among functions merit a name: We say that the functions f, f 2,, f m are 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, c 2,, c m not all zero such that c f (t) + c 2 f 2 (t) + + c m f m (t) = (7) for all t in I A set of functions that is not linearly dependent is said to be linearly independent This means that a sum of the form (7) implies that c = c 2 = = c m = Example 3 Show that the functions f (t) = e t, f 2 (t) = e 2t, and f 3 (t) = 3e t 2e 2t are linearly dependent on (, ) 36

37 Solution Since f 3 (t) = 3f (t) 2f 2 (t), the given functions are linearly dependent on (, ) The concept of linear independence can be used to test a fundamental set of solutions to the equation y (n) + p n (t)y (n ) + + p (t)y + p (t)y = (8) Theorem 3 The solution set {y, y 2,, y n } is a fundamental set of solutions to y (n) + p n (t)y (n ) + + p (t)y + p (t)y = where p n (t),, p (t), p (t) are continuous functions in a < t < b if and only if the functions y, y 2,, y n are linearly independent Proof Suppose first that {y, y 2,, y n } is a fundamental set of solutions Then by Theorem 285 there is a < t < b such that W (t ) Suppose that c y (t) + c 2 y 2 (t) + + c n y n (t) = for all a < t < b By repeated differentiation of the previous equation we find c y (t) + c 2 y 2(t) + + c n y n(t) = c y (t) + c 2 y 2(t) + + c n y n(t) = c y (n ) (t) + c 2 y (n ) 2 (t) + + c n y n (n ) (t) = Thus, one finds c, c 2,, c n by solving the system c y (t ) + c 2 y 2 (t ) + + c n y n (t ) = c y (t ) + c 2 y 2(t ) + + c n y n(t ) = c y (t ) + c 2 y 2(t ) + + c n y n(t ) = c y (n ) (t ) + c 2 y (n ) 2 (t ) + + c n y n (n ) (t ) = Solving this system using Cramer s rule one finds c = c 2 =, c n = W (t ) = 37

38 Thus, y (t), y 2 (t),, y n (t) are linearly independent Conversely, suppose that {y, y 2,, y n } is a linearly independent set Suppose that {y, y 2,, y n } is not a fundamental set of solutions Then by Theorem 32, W (t) = for all a < t < b Choose any a < t < b Then W (t ) = But this says that the matrix y (t ) y 2 (t ) y n (t ) y (t ) y 2(t ) y n(t ) y (n ) (t ) y (n ) 2 (t ) y n (n ) (t ) is not invertible which means that there exist c, c 2,, c n not all zero such that c y (t ) + c 2 y 2 (t ) + + c n y n (t ) = c y (t ) + c 2 y 2(t ) + + c n y n(t ) = c y (t ) + c 2 y 2(t ) + + c n y n(t ) = c y (n ) (t ) + c 2 y (n ) 2 (t ) + + c n y n (n ) (t ) = Now, let y(t) = c y (t)+c 2 y 2 (t)+ +c n y n (t) for all a < t < b Then y(t) is a solution to the differential equation and y(t ) = y (t ) = = y (n ) (t ) = But the zero function also is a solution to the initial value problem By the existence and uniqueness theorem (ie, Theorem 29) we must have c y (t) + c 2 y 2 (t) + + c n y n (t) = for all a < t < b with c, c 2,, c n not all equal to But this means that y, y 2,, y n are linearly depedent which contradicts our assumption that y, y 2,, y n are linearly independent Remark 3 The fact that {y, y 2,, y n } are solutions is very critical That is, if y, y 2,, y n are merely differentiable functions then it is possible for them to be linearly independent and yet have a vanishing Wronskian See Section 7 Next, we will show how to generate new fundamental sets from a given one and therefore establishing the fact that a linear homogeneous differential equation has many fundamental sets of solutions We also show how different fundamental sets are related to each other For this, let us start with a fundamental set {y, y 2,, y n } of solutions to (8) If {y, y 2,, y n } are n solutions then they can be written as linear combinations of the {y, y 2,, y n } 38

39 That is, or in matrix form as a y + a 2 y a n y n = y a 2 y + a 22 y a n2 y n = y 2 a n y + a 2n y a nn y n = y n [ [ y y 2 y n = y y 2 y n a a 2 a 3 a n a 2 a 22 a 23 a 2n a n a n2 a n3 a nn Theorem 32 {y, y 2,, y n } is a fundamental set if and only if det(a) where A is the coefficient matrix of the above matrix equation Proof By differentiating (n-) times the system one can easily check that y y 2 y n y y 2 y n a y + a 2 y a n y n = y a 2 y + a 22 y a n2 y n = y 2 a n y + a 2n y a nn y n = y n y (n ) y (n ) 2 y (n ) n = y y 2 y n y y 2 y n y (n ) y (n ) 2 y n (n ) a a 2 a n a 2 a 22 a 2n a n a n2 a nn By taking the determinant of both sides and using the fact that the determinant of a product is the product of determinants then we can write W (t) = det(a)w (t) Since W (t), W (t) (ie, {y, y 2,, y n } is a fundamental set) if and only if det(a) 39

40 Example 32 The set {y (t), y 2 (t), y 3 (t)} = {, e t, e t } is fundamental set of solutions to the differential equation y y = (a) Show that {y (t), y 2 (t), y 3 (t)} = {cosh t, sinh t, 2+sinh t} is a solution set (b) Determine the coefficient matrix A described in the previous theorem (c) Determine whether this set is a fundamental set by calculating the determinant of the matrix A Solution (a) Since y = sinh t, y = cosh t, and y (t) = sinh t we have y y = so that y is a solution A similar argument holds for y 2 and y 3 (a) Since y (t) = + 2 et + 2 e t, y 2 (t) = 2 et + 2 e t, y 3 (t) = et 2 e t we have A = 2 /2 /2 /2 /2 /2 /2 (c) One can easily find that det(a) = 3 2 so that {y (t), y 2 (t), y 3 (t)} is a fundemantal set of solutions 4

41 Practice Problems Problem 3 Determine if the following functions are linearly independent y (t) = e 2t, y 2 (t) = sin (3t), y 3 (t) = cos t Problem 32 Determine whether the three functions : f(t) = 2, g(t) = sin 2 t, h(t) = cos 2 t, are linearly dependent or independent on < t < Problem 33 Determine whether the functions, y (t) = ; y 2 (t) = + t; y 3 (t) = + t + t 2 ; are linearly dependent or independent Show your work Problem 34 Consider the set of functions {y (t), y 2 (t), y 3 (t)} = {t 2 +2t, αt+, t+α} For what value(s) α is the given set linearly depedent on the interval < t <? Problem 35 Determine whether the set {y (t), y 2 (t), y 3 (t)} = {t t +, t 2, t} is linearly independent or linearly dependent on the given interval (a) t < (b) < t (c) < t < In Problems 36-37, for each differential equation, the corresponding set of functions {y (t), y 2 (t), y 3 (t)} is a fundamental set of solutions (a) Determine whether the given set {y (t), y 2 (t), y 3 (t)} is a solution set to the differential equation (b) If {y (t), y 2 (t), y 3 (t)} is a solution set then find the coefficient matrix A such that y y 2 y 3 = a a 2 a 3 a 2 a 22 a 23 a 3 a 32 a 33 (c) If {y (t), y 2 (t), y 3 (t)} is a solution set, determine whether it is a fundamental set by calculating the determinant of A y y 2 y 3 4

42 Problem 36 y + y = {y (t), y 2 (t), y 3 (t)} = {, t, e t } {y (t), y 2 (t), y 3 (t)} = { 2t, t + 2, e (t+2) } Problem 37 t 2 y + ty y =, t > {y (t), y 2 (t), y 3 (t)} = {t, ln t, t 2 } {y (t), y 2 (t), y 3 (t)} = {2t 2, 3, ln (t 3 )} 42

43 32 Higher Order Homogeneous Linear Equations with Constant Coefficients In this section we investigate how to solve the nth order linear homogeneous equation with constant coefficients The general solution is given by y (n) + a n y (n ) + + a y + a = (9) y(t) = c y + c 2 y c n y n where {y, y 2,, y n } is a fundamental set of solutions What was done for second-order, linear homogeneous equations with constant coefficients holds, with the obvious modifications, for higher order analogs As for the second order case, we seek solutions of the form y(t) = e rt, where r is a constant (real or complex-valued) to be found Inserting into (9) we find (r n + a n r n + a r + a )e rt = We call P (r) = r n + a n r n + a r + a the characteristic polynomial and the equation r n + a n r n + a r + a = () the characteristic equation Thus, for y(t) = e rt to be a solution to (9) r must satisfy () Example 32 Solve: y 4y + y + 6y = Solution The characteristic equation is r 3 4r 2 + r + 6 = We can factor to find the roots of the equation A calculator can efficiently do this, or you can use the rational root theorem to get (r + )(r 2)(r 3) = 43

44 Thus, the roots are The Wronskian r =, r = 2, r = 3 e t e 2t e 3t e t 2e 2t 3e 3t e t 4e 2t 9e 3t = 2e4t Hence, {e t, e 2t, e 3t } is a fundamental set of solutions and the general solution is y = c e t + c 2 e 2t + c 3 e 3t In the previous example, the characteristic solution had three distinct roots and the corresponding set of solutions formed a fundamental set This is always true according to the following theorem Theorem 32 Assume that the characteristic equation r n + a n r n + a r + a = has n distinct roots r, r 2,, r n (real valued or complex valued) Then the set of solutions {e r t, e r 2t,, e rnt } is a fundamental set of solution to the equation y (n) + a n y (n ) + + a y + a = Proof For a fixed number t we consider the Wronskian W (t ) = e r t e r 2t e rnt r e r t r 2 e r 2t r n e rnt re 2 r t r2e 2 r 2t rne 2 rnt r n e r t r2 n e r 2t rn n e rnt Now, in linear algebra one proves that if a row or a column of a matrix is multiplied by a constant then the determinant of the new matrix is the determinant of the old matrix multiplied by that constant It follows that W (t) = e r t e r 2t e rnt 44 r r 2 r n r 2 r2 2 rn 2 r n r2 n rn n

45 The resulting determinant above is the well-known Vandermonde determinant Its value is the product of all factors of the form r j r i where j > i Since r j r i for i j, this determinant is not zero and consequently W (t ) This establishes that {e r t, e r 2t,, e rnt } is a fundamental set of solutions Next, we consider characteristic equations whose roots are not all distinct For example, if α is a real root that appears k times (in this case we say that α is a root of multiplicity k), that is, P (r) = (r α) k q(r), where q(α), then the k linearly independent solutions are given by e αt, te αt, t 2 e αt,, t k e αt The remaining n k solutions needed to complete the fundamental set of solutions are determined by examining the roots of q(r) = If, on the other hand, α±iβ are conjugate complex roots each of multiplicity k, that is P (r) = (r r ) k (r r ) k p(r) where r = α + iβ and p(r ), p(r ) then the 2k linearly independent solutions are given by and Example 322 Find the solution to e αt cos βt, te αt cos βt,, t k e αt cos βt e αt sin βt, te αt sin βt,, t k e αt sin βt y (5) + 4y =, y() = 2, y () = 3, y () =, y () =, y(4)() = Solution We have the characteristic equation r 5 + 4r 3 = r 3 (r 2 + 4) = Which has a root of multiplicity 3 at r = and complex roots r = 2i and r = 2i We use what we have learned about repeated roots and complex roots to get the general solution Since the multiplicity of the repeated root is 3, we have y (t) =, y 2 (t) = t, y 3 (t) = t 2 The complex roots give the other two solutions 45

46 The general solution is y 4 (t) = cos (2t) and y 5 (t) = sin (2t) y(t) = c + c 2 t + c 3 t 2 + c 4 cos (2t) + c 5 sin (2t) Now Find the first four derivatives y (t) = c 2 + 2c 3 t 2c 4 sin (2t) + 2c 5 cos (2t) y (t) = 2c 3 4c 4 cos (2t) 4c 5 sin (2t) y (t) = 8c 4 sin (2t) 8c 5 cos (2t) y(4)(t) = 6c 4 cos (2t) + 6c 5 sin (2t) Next plug in the initial conditions to get Solving these equations we find The solution is 2 = c + c 4 3 = c 2 + 2c 5 = 2c 3 4c 4 = 8c 5 = 6c 4 c = 3/6, c 2 = 23/4, c 3 = 5/8, c 4 = /6, c 5 = /8 y(t) = t t2 + 6 cos (2t) + sin (2t) 8 Solving the Equation y (n) ay = The characteristic equation corresponding to the differential equation y (n) ay = is r n a = The fundamental theorem of algebra asserts the existence of exactly n roots (real or complex-valued) To find these roots, we write a in polar form a = a e iα where α = if a > and α = π is a < (since e iπ = cos π + i sin π = ) Also, since e i2kπ = for any integer k then we can write a = a e (α+2kπ)i Thus, the charactersitic equation is r n = a e (α+2kπ)i 46

47 Taking the nth root of both sides we find r = a n e (α+2kπ)i n The n distinct roots are generated by taking k =,, 2,, n We illustrate this in the next example Example 323 Find the general solution of y (6) + y = Solution In this case the characteristic equation is r 6 + = or r 6 = = e i(2k+)π Thus, r = e i (2k+)π 6 where k is an integer Replacing k by,,2,3,4,5 we find Thus, the general solution is r = 3 + i 2 2 r = i r 2 = 3 + i 2 2 r 3 = 3 i 2 2 r 4 = i r 5 = 3 i y(t) = c e 2 t cos t 2 +c 3 2 2e t sin t 2 +c 3e 3 2 t cos t 2 +c 3 2 4e t sin t 2 +c 5 cos t+c 6 sin t 47

48 Practice Problems Problem 32 Solve y + y y y = Problem 322 Find the general solution of 6y (4) 8y + y = Problem 323 Solve the following constant coefficient differential equation : Problem 324 Solve y (4) 6y = Problem 325 Solve the initial-value problem y y = y + 3y + 3y + y =, y() =, y () =, y () = Problem 326 Given that r = is a solution of r 3 + 3r 2 4 =, find the general solution to y + 3y 4y = Problem 327 Given that y (t) = e 2t is a solution to the homogeneous equation, find the general solution to the differential equation y 2y + y 2y = Problem 328 Suppose that y(t) = c cos t + c 2 sin t + c 3 cos (2t) + c 4 sin (2t) is the general solution to the equation Find the constants a, a, a 2, and a 3 y (4) + a 3 y + a 2 y + a y + a y = 48

49 Problem 329 Suppose that y(t) = c + c 2 t + c 3 cos 3t + c 4 sin 3t is the general solution to the homogeneous equation y (4) + a 3 y + a 2 y + a y + a y = Determine the values of a, a, a 2, and a 3 Problem 32 Suppose that y(t) = c e t sin t+c 2 e t cos t+c 3 e t sin t+c 4 e t cos t is the general solution to the homogeneous equation y (4) + a 3 y + a 2 y + a y + a y = Determine the values of a, a, a 2, and a 3 Problem 32 Consider the homogeneous equation with constant coefficients y (n) + a n y (n ) + + a y + a = Suppose that y (t) = t, y 2 (t) = e t, y 3 (t) = cos t are several functions belonging to a fundamental set of solutions to this equation What is the smallest value for n for which the given functions can belong to such a fundamental set? What is the fundamemtal set? Problem 322 Consider the homogeneous equation with constant coefficients y (n) + a n y (n ) + + a y + a = Suppose that y (t) = t 2 sin t, y 2 (t) = e t sin t are several functions belonging to a fundamental set of solutions to this equation What is the smallest value for n for which the given functions can belong to such a fundamental set? What is the fundamemtal set? Problem 323 Consider the homogeneous equation with constant coefficients y (n) + a n y (n ) + + a y + a = Suppose that y (t) = t 2, y 2 (t) = e 2t are several functions belonging to a fundamental set of solutions to this equation What is the smallest value for n for which the given functions can belong to such a fundamental set? What is the fundamemtal set? 49

50 33 Non Homogeneous nth Order Linear Differential Equations We consider again the nth order linear nonhomogeneous equation y (n) + p n (t)y (n ) + + p (t)y + p (t)y = g(t) () where the functions p, p,, p n and g(t) are continuous functions for a < t < b The solution structure established for second order linear nonhomogeneous equations applies as well in the nth order case Theorem 33 Let {y (t), y 2 (t),, y n (t)} be a fundamental set of solutions to the homogeneous equation y (n) + p n (t)y (n ) + + p (t)y + p (t)y = and y p (t) be a particular solution of the nonhomogeneous equation y (n) + p n (t)y (n ) + + p (t)y + p (t)y = g(t) The general solution of the nonhomogeneous equation is given by y(t) = y p (t) + c y (t) + c 2 y 2 (t) + + c n y n (t) Proof Let y(t) be any solution to equation () Since y p (t) is also a solution, we have (y y p ) (n) + p n (t)(y y p ) (n ) + + p (t)(y y p ) + p (t)(y y p ) = y (n) +p n (t)y (n ) + +p (t)y +p (t)y (y p (n) +p n (t)y p (n ) + +p (t)y p+p (t)y p ) = g(t) g(t) = Therefore y y p is a solution to the homogeneous equation But {y, y 2,, y n } is a fundamental set of solutions to the homogeneous equation so that there exist unique constants c, c 2,, c n such that y(t) y p (t) = c y (t)+c 2 y 2 (t)+ + c n y n (t) Hence, y(t) = y p (t) + c y (t) + c 2 y 2 (t) + + c n y n (t) 5

51 Since the sum c y (t) + c 2 y 2 (t) + + c n y n (t) represents the general solution to the homogeneous equation, we will denote it by y h so that the general solution of () takes the form y(t) = y h (t) + y p (t) It follows from the above theorem that finding the general solution to nonhomogeneous equations consists of three steps: Find the general solution y h of the associated homogeneous equation 2 Find a single solution y p of the original equation 3 Add together the solutions found in steps and 2 The superposition of solutions is valid only for homogeneous equations and not true in general for nonhomogeneous equations (Recall the case n = 2 in Section 22) However, we can have a property of superposition of nonhomogeneous if one is adding two solutions of two different nonhomogeneous equations More precisely, we have Theorem 332 Let y (t) be a solution of y (n) + p n (t)y (n ) + + p (t)y + p (t)y = g (t) and y 2 (t) a solution of y (n) + p n (t)y (n ) + + p (t)y + p (t)y = g 2 (t) Then for any constants c and c 2 the function Y (t) = c y (t) + c 2 y 2 (t) is a solution of the equation Proof We have y (n) + p n (t)y (n ) + + p (t)y + p (t)y = c g (t) + c 2 g 2 (t) L[Y = c (y (n) + p n (t)y (n ) + + p (t)y + p (t)y ) + c 2 (y (n) 2 + p n (t)y (n ) p (t)y 2 + p (t)y 2 ) = c g (t) + c 2 g 2 (t) Next, we discuss methods for determining y p (t) The techinque we discuss first is known as the method of undetermined coefficients This method requires that we make an initial assumption about the form of the particular solution y p (t), but with the coefficients left unspecified, thus the name of the method We then substitute the assumed expression into equation () and attempt to determine the coefficients as to satisfy that 5