Question

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

Answer 1

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.