#2y'' - 2y' - 5y = 0, qquad y(0) = 2, qquad y'(0) = 3#
#y'' = y' + 5/2 y#
#((y''),(y')) = ((1, 5/2),(1,0)) ((y'),(y))#
#bb y^' = M bb y #
Which solves trivially as:
Using the exponential definition:
#= ( bb I + M y + M^2 ( y)^2/(2!) + ... ) bby_0#
The eigenvalues of #M# are:
- #lambda_("1,2") = 1/2 - sqrt(11)/2 , 1/2 + sqrt(11)/2 #
With eigenvectors:
- #bb alpha_("1,2") = ((1/2 - sqrt(11)/2) ,( 1)), qquad ((1/2 + sqrt(11)/2 ),( 1))#
Because #M# delivers 2 eigenvalues, it can be diagonalised:
- #M = CDC^(-1) = C ((lambda_1, 0),(0, lambda_2)) C^(-1)#
#bb y = ( C bb I C^(-1) + CDC^(-1) x + CD^2C^(-1) x^2/(2!) ... ) bby_0#
#= C( bb I + ((lambda_1, 0),(0, lambda_2)) x + ((lambda_1^2, 0),(0, lambda_2^2)) x^2/(2!) + ... ) C^(-1) bby_0#
#= C( (1 + lambda_1 x + (lambda_1 x)^2/(2!) + ...,0),(0,1 + lambda_2 x + (lambda_2 x)^2/(2!) + ...)) C^(-1) bby_0#
#= C( (e^(lambda_1 x),0),(0,e^(lambda_2 x) ) ) C^(-1) ((3), (2))#
#C = ((lambda_1, lambda_2),(1, 1))#
#C^(-1) =1/(lambda_1 - lambda_2) ((1, - lambda_2),(- 1, lambda_1))#
#implies bby = 1/(lambda_1 - lambda_2) ((lambda_1, lambda_2),(1, 1)) ( (e^(lambda_1 x),0),(0,e^(lambda_2 x) ) ) ((1, - lambda_2),(- 1, lambda_1)) ((3), (2))#
Doing the products from the end (as this simplifies to a 2 x 1 matrix):
# = 1/(lambda_1 - lambda_2) ((lambda_1, lambda_2),(1, 1)) ( (e^(lambda_1 x),0),(0,e^(lambda_2 x) ) ) ((3- 2 lambda_2), (- 3 +2 lambda_1))#
# = 1/(lambda_1 - lambda_2) ((lambda_1, lambda_2),(1, 1))((e^(lambda_1 x)( 3- 2 lambda_2)), (e^(lambda_2 x) (- 3 +2 lambda_1)))#
# = 1/(lambda_1 - lambda_2)((lambda_1 e^(lambda_1 x)( 3- 2 lambda_2) + lambda_2 e^(lambda_2 x) (- 3 +2 lambda_1)), (e^(lambda_1 x)( 3- 2 lambda_2) + e^(lambda_2 x) (- 3 +2 lambda_1)))#
From the bottom line of #bb y = ((y'),(y))#:
#y = 1/(sqrt11) (e^((1/2(1 - sqrt(11))) x)( 3- 2 (1/2 + sqrt(11)/2)) + e^((1/2(1 + sqrt(11))) x) (- 3 +2 (1/2 - sqrt(11)/2)))#
