JUST THE MATHS SLIDES NUMBER 15.10 ORDINARY DIFFERENTIAL - PDF document

“JUST THE MATHS” SLIDES NUMBER 15.10 ORDINARY DIFFERENTIAL EQUATIONS 10 (Simultaneous equations (C)) by A.J.Hobson 15.10.1 Matrix methods for non-homogeneous systems

UNIT 15.10 - ORDINARY DIFFERENTIAL EQUATIONS 10 SIMULTANEOUS EQUATIONS (C) 15.10.1 MATRIX METHODS FOR NON-HOMOGENEOUS SYSTEMS The general solution of a single linear differential equa- tion with constant coefficients is made up of a particular integral and a complementary function. The complementary function is the general solution of the corresponding homogeneous differential equation. A similar principle is now applied to a pair of simultane- ous non-homogeneous differential equations of the form d x 1 = ax 1 + bx 2 + f ( t ) , d t d x 2 = cx 1 + dx 2 + g ( t ) . d t The method will be illustrated by an example. 1

EXAMPLE Determine the general solution of the simultaneous dif- ferential equations d x 1 = x 2 , − − − − − − − − − − − (1) d t d x 2 = − 4 x 1 − 5 x 2 + g ( t ) , − − − − (2) d t where g ( t ) is (a) t , (b) e 2 t (c) sin t , (d) e − t . Solutions (i) First, we write the differential equations in matrix form as d  0 1  0  x 1  x 1          =  +  .  g ( t ) .         x 2 − 4 − 5 x 2 1 d t This may be interpreted as dX dt = MX + N g ( t ) , where  x 1  0 1  0        and N =  , M =  . X =       x 2 − 4 − 5 1 2

(ii) Secondly, we consider the corresponding homogeneous system dX dt = MX . The characteristic equation is 0 − λ 1 � � � � � = 0 � � � � � − 4 − 5 − λ � � � � and gives λ (5 + λ ) + 4 = 0 , or λ 2 + 5 λ + 4 = 0 , or ( λ + 1)( λ + 4) = 0 . Hence, λ = − 1 or λ = − 4. 3

(iii) The eigenvectors of M are obtained from the homo- geneous equations − λk 1 + k 2 = 0 , − 4 k 1 − (5 + λ ) k 2 = 0 . Hence, when λ = − 1, we solve k 1 + k 2 = 0 , − 4 k 1 − 4 k 2 = 0 . These are satisfied by any two numbers in the ratio k 1 : k 2 = 1 : − 1. Also, when λ = − 4, we solve 4 k 1 + k 2 = 0 , − 4 k 1 − k 2 = 0 . These are satisfied by any two numbers in the ratio k 1 : k 2 = 1 : − 4. 4

The complementary function may now be written in the form  1  1      e − t + B  e − 4 t , A     − 1 − 4 where A and B are arbitrary constants. (iv) In order to obtain a particular integral for the equa- tion dX dt = MX + N g ( t ) , we note the second term on the right hand side and in- vestigate a trial solution of a similar form. The three cases in this example are as follows: (a) g ( t ) ≡ t Trial solution X = P + Q t, where P and Q are constant matrices of order 2 × 1. We require that Q = M(P + Q t ) + N t. 5

Equating the matrix coefficients of t and the constant matrices, MQ + N = 0 and Q = MP . Thus, Q = − M − 1 N and P = M − 1 Q . Using M − 1 = 1  − 5 − 1    ,   4 0 4 we obtain Q = − 1  − 5 − 1  0  0 . 25        =  .       4 0 1 0  4 and P = 1  − 5 − 1  0 . 25  − 0 . 3125        =  .  .       4 0 0 0 . 25 4 The general solution, in this case, is  1  1  − 0 . 3125  0 . 25          e − t + B  e − 4 t +  +  t. X = A         − 1 − 4 0 . 25 0 6

(b) g ( t ) ≡ e 2 t Trial solution X = P e 2 t . We require that 2P e 2 t = MP e 2 t + N e 2 t . That is, 2P = MP + N . The matrix P may now be determined from the formula (2I − M)P = N . In more detail,  2 − 1    . P = N .   4 7 Hence, P = 1  = 1  7 1  1  7        .  .       − 4 2 0 − 4 18 18 7

The general solution, in this case, is  e − 4 t + 1  1  1  7        e − t + B  e 2 t . X = A       − 1 − 4 − 4 18 (c) g ( t ) ≡ sin t Trial solution X = P sin t + Q cos t. We require that P cos t − Q sin t = M(P sin t + Q cos t ) + N sin t. Equating the matrix coefficients of cos t and sin t , P = MQ and − Q = MP + N . This means that − Q = M 2 Q + N or (M 2 + I)Q = − N . Thus, Q = − (M 2 + I) − 1 N .  0 1  0 1  1 0  − 3 − 5         M 2 + I =  +  =  .         − 4 − 5 − 4 − 5 0 1 20 22  8

Hence, Q = − 1  = 1  22 5  0  − 5        .       − 20 − 3 1 3  34 34 Also, P = MQ = 1  = 1  0 1  − 5  3        .  .       − 4 − 5 3 5 34 34 The general solution, in this case, is  e − 4 t + 1  sin t + 1  1  1  3  − 5          e − t + B  cos t. X = A         − 1 − 4 5 3 34 34 (d) g ( t ) ≡ e − t In this case, the function, g ( t ), is already included in the complementary function and it becomes necessary to assume a particular integral of the form X = (P + Q t ) e − t , where P and Q are constant matrices of order 2 × 1. We require that Q e − t − (P + Q t ) e − t = M(P + Q t ) e − t + N e − t . 9

Equating the matrix coefficients of te − t and e − t , we obtain − Q = MQ and Q − P = MP + N . The first of these conditions shows that Q is an eigenvec- tor of the matrix M corresponding the eigenvalue − 1. From earlier work,  1    , Q = k   − 1 for any constant k . Also, (M + I)P = Q − N; or, in more detail,  1 1  1  0  p 1          = k  .  .  −         − 4 − 4 − 1 1 p 2 10

Hence, p 1 + p 2 = k, − 4 p 1 − 4 p 2 = − k − 1 . Thus, k = k +1 giving k = 1 3 ; and the matrix P is given 4 by  = 1  0  1 l        + l  , P =       1 1 − 1  3 − l 3 for any number, l . Taking l = 0 for simplicity, a particular integral is, there- fore, X = 1  0  1        e − t .   +   t       1 − 1 3    The general solution is  e − 4 t + 1  1  1  0  1            e − t + B  e − t .   +   t X = A           − 1 − 4 1 − 1 3    11

Note: In examples for which neither f ( t ) nor g ( t ) is identically equal to zero, the particular integral may be found by adding together the separate forms of particular integral for f ( t ) and g ( t ) and writing the system of differential equations in the form dX dt = MX + N 1 f ( t ) + N 2 g ( t ) , where  1  0      and N 2 =  . N 1 =     0 1 For instance, if f ( t ) ≡ t and g ( t ) ≡ e 2 t , the particular integral would take the form X = P + Q t + R e 2 t , where P, Q and R are matrices of order 2 × 1. 12