1) Obtain the general solution(s) for the system:

dx/dt = 2x + y

dy/dt = 2x + 3y

* Solution*:

In this case, the alert math maven will note the x, y derivatives are with respect to t, but the equivalence is to x, y forms with no t variable present. We reconcile our terms and allow the solution to proceed via the following substitutions:

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 + l

^{2}– 2 =l

^{2}– 5l + 4 = 0Or: (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)

2)Obtain the general solution(s) for the system:

dx1/ dt = 3x1 - 4x2

dx2/dt = x1 - x2

When roots are real and equal, then assume soln. of form:

x1 = A exp(lt) and x2 = B exp(lt)

If Det (A - lI) D = 0 only has one soln. l then (A - lI) D = 0 has at least one

*independent solution*. A first solution is: X1 = D exp (lt)Now take: X = D exp (lt) + Bt exp (lt)

dX/ dt = l D exp (lt) + B exp(lt) + l Bt exp(lt)

= AX + AD exp (lt) + ABt exp (lt)

For all t we need to satisfy:

B = (A - lI) D + t (A - lI) B

Take (A - lI) B = 0

(A - lI) D = B

The independent solution is then:

X2 = D exp (lt) + Bt exp (lt)

Applied to the given linear system we get:

The characteristic equation is then:

l

^{2}– 2l + 1 = 0Whence:

2d1 – 4d2 = 0

d1 - 2d2 = 0

=> 3d1 - 6d2 = 0

Or: d1 - 2d2 = 0

Then:

d1 = 2d2 so: d1 = 2 and d2 = 1

Eigenvector =

(2)

(1)

Finally:

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