Question

How to determine the general solution to the system of differential equations ? `y'_1(t) = 9y_1(t) - y_2(t)` `y'_2 (t) = y_1(t) + 7y_2(t)`

Answer 1

`y_1 = Ae^(8t)+B(t+1)e^(8t)`
`y_2 = Ae^(8t)+Bte^(8t)`

Explanation:

We have:

`y_1' = 9y_1 - y_2`
`y_2' = y_1+7 y_2`

Which 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) ) = ( (9, -1), (1, 7) ) ( (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`
`:. | (9-lamda, -1), (1, 7-lamda) |`
`:. (9-lamda)(7-lamda) - (1)(-1) = 0`
`:. 63 -9lamda-7lamda+lamda^2+1 = 0`
`:. lamda^2-16lamda+64 = 0`
`:. (lamda -8)^2`
`:. lamda_1,lamda_2 = 8`

Thus, we have a repeated eigenvalue. Next we seek the corresponding eigenvector, 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, lamda_2 = 8`

`( (9-8, -1), (1, 7-8) ) ( (u_1), (u_2) ) = ( (0), (0) )`
`:. ( (1, -1), (1, -1) ) ( (u_1), (u_2) ) = ( (0), (0) )`
`:. u_1-u_2 = 0`
Put `u_1 = 1 => u_2=1 => bb(ul(e_1)) = ( (1), (1) )`

Hence we have found the eigenvalues:

`lamda_1,lamda_2 = 8 \ \ \` (repeated)

And corresponding eigenvectors:

`bb(ul(e_1)) = ( (1), (1) )`

As we have a repeated eigenvalue, we can only gain one independent solution and we have a defective system, and thus we can find a second independent solution by solving:

`(bb(A)- lamda_1 bb(I)) bb(ul(u_2)) = bb(ul(e_1))`

`:. (bb(A)- 8 bb(I)) bb(ul(u_2)) = bb(ul(e_1))`

`:. ( (1, -1), (1, -1) ) ( (u_1), (u_2) ) = ( (1), (1) ) =>`

`:. u_1 - u_2 = 1 => u_1=1-u_2`

Arbitrarily put `u_2=0 => u_1=1`

So we can gain a second independent solution using

`bb(ul(e_2)) = ( (1), (0) )`

And so the solution of the coupled system is:

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

i.e. it is

`( (y_1), (y_2) ) = Ae^(8t)( (1), (1) ) + Be^(8t)(t( (1), (1) )+( (1), (0) ))`

Or in non-matrix form:

`y_1 = Ae^(8t)+B(t+1)e^(8t)`
`y_2 = Ae^(8t)+Bte^(8t)`