(constant coefficient systems phase plane method) # y_1'=y_1+y_2 , y_2'=3y_1-y_2 #?

1 Answer
Apr 9, 2018

Assuming the independent variable is #t#:

# y_1 = \ \ \ \ \ \ \Ae^(-2t) + Be^(2t) #
# y_2 = -3Ae^(-2t) + Be^(2t) #

Explanation:

#y_1'=y_1 + y_2 #
# y_2' = 3y_1-y_2 #

Represents a coupled set of First Order Linear differential equation. As with most topics in Linear Mathematics, we can represent this in matrix form as follows:

# ( (y_1'), (y_2) ) = ( (1, 1), (3, -1) ) ( (y_1), (y_2) ) #

Or in shorthand:

# bb(ul(y)') = bb(A) \ bb(ul(y)) #

In order to solve this coupled system, we seek the eigenvalues of the matrix #bb(A)#, whose characteristic equation is given by:

# det( bb(A)-lamda bb( I) ) = 0 #
# :. | (1-lamda, 1), (3, -1-lamda) | #
# :. (1-lamda)(-1-lamda) - (1)(3) = 0 #
# :. -1 -lamda+lamda+lamda^2-3 = 0 #
# :. lamda^2-4 = 0 #
# :. lamda = +- 2 #

Next we seek the corresponding eigenvectors, which by definition will satisfy:

# bb(A) \ bb(ul(u)) = lamda bb(ul(u)) => (bb(A)-lamda bb(I)) \ bb(ul(u)) = 0#

Consider #lamda_1 = -2#

# ( (1+2, 1), (3, -1+2) ) ( (u_1), (u_2) ) = ( (0), (0) )#
# :. ( (3, 1), (3, 1) ) ( (u_1), (u_2) ) = ( (0), (0) )#
# :. 3u_1+u_2 = 0#
# :. u_1 = 1; u_2=-3#, say #=> bb(ul(e_1)) = ( (1), (-3) ) #

Consider #lamda_1 = 2#

# ( (1-2, 1), (3, -1-2) ) ( (u_1), (u_2) ) = ( (0), (0) )#
# :. ( (-1, 1), (3, -3) ) ( (u_1), (u_2) ) = ( (0), (0) )#
# :. -u_1+u_2=0#
# :. u_1 = 1; u_2=1#, say #=> bb(ul(e_2)) = ( (1), (1) ) #

Hence we have found the eigenvalues:

# lamda_1 = -2 \ \ \ ; lamda_2 = 2 #

And corresponding eigenvectors:

# bb(ul(e_1)) = ( (1), (-3) ) \ \ \ ; bb(ul(e_2)) = ( (1), (1) ) #

And so the solution of the coupled system is:

# ( (x), (y) ) = Ae^(lamda_1t)bb(ul(e_1)) + Be^(lamda_1t)bb(ul(e_2)) #

i.e. it is

# ( (y_1), (y_2) ) = Ae^(-2t)( (1), (-3) ) + Be^(2t) ( (1), (1) ) #

Or in non-matrix form:

# y_1 = \ \ \ \ \ \ \Ae^(-2t) + Be^(2t) #
# y_2 = -3Ae^(-2t) + Be^(2t) #

Validation:

Differentiating wrt #t#

# y_1' = -2Ae^(-2t) + 2Be^(2t) #
# y_2' = 6Ae^(-2t) + 2Be^(2t) #

And substituting:

# y_1 + y_2 = Ae^(-2t) + Be^(2t) + -3Ae^(-2t) + Be^(2t)#
# " " = -2Ae^(-2t) + 2Be^(2t) #
# " " = y _1' #

# 3y_1-y_2 = 3Ae^(-2t) + 3Be^(2t) +3Ae^(-2t) - Be^(2t) #
# " " = 6Ae^(-2t) + 2Be^(2t)#
# " " = y_2'#

Confirming the solution is sound.