Solution to Problem 1 of June 30 Set for Linear, Homogeneous DE Systems

We need to find the general solution(s) for the system:

dx/dt = 2x + y

dy/dt = 2x + 3y

 

This problem is used to illustrate a slightly different method of solution, as the end procedure is in the same mode as earlier problems. In this case, the alert blog readers and ultra math mavens will note the x, y derivatives are with respect to t, but the equivalence is to x, y forms with no t variable present.

How do we reconcile our terms and allow the solution to proceed? Well, we use (since we expect the end forms of our solutions to be in exp form anyway):

x = A exp(lt) and y = B exp(lt)

With these substitutions we than re-write our system in a consistent form:

l A exp(lt)   =  2 A exp(lt)    + B exp(lt)

lB exp(lt)   =  2 A exp(lt)    + 3B exp(lt)

(Recall: dx/dt = d (A exp(lt))/dt = l A exp(lt) - Analogous for dy/dt!)

Now, collect like terms:

2 A exp(lt) - l A exp(lt)       + B exp(lt) =  0

2 A exp(lt)    + 3B exp(lt)   -lB exp(lt)   = 0

Factor out the exp(lt) terms top and bottom to get:

(2 - l) A  + B = 0

2A  + (3 - l) B = 0

Set up the determinant as per prior problems: (A - l) D = 0 =

(2 - l…………1)
(2………...3 - l )

Find the characteristic equation, viz.

(2 - l) (3 - l) – 2 = 0 = 6 - 3 l - 2 l + l² – 2 =

l² – 5l  + 4 = 0

Or: (l - 4) (l - 1) = 0

So we have eigenvalues : l1 = 4  and l2 = 1

As before with previous problems, substitute the eigenvalue l1 = 4   into the matrix to get:

[-2 ……..1] [A]

[2……..-1] [B]


Whence: -2 A + B = 0 and 2A – B = 0


So, A = 1, B = 2, so c1 =

[1]
[2]

The solution is then: X1 = c1 exp (4t)

   

Now, substitute the eigenvalue l2 = 1   into the matrix to get:

  [1 ……..1] [A]

  [2……..2] [B]

Whence:  A + B = 0 and 2A + 2B = 0 so: A = 1 and B = -1

so c2 =

[1]
[-1]

The solution is then: X2 = c2 exp (t)

The full general solution for the system is:

X = X1 + X2 = c1 exp (4t) +  c2 exp (t)