dx/dt = 3x + 2y

dy/dt = -5x + y

Much of the procedure for solution of this system is the
same as for the earlier examples. Thus, assume a solution of the form:

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

Now, substitute the preceding into the system to get:

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

lB
exp(lt) =
-5A exp(lt) + B
exp(lt)

This leads to:

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

-5 A exp(lt) + B exp(lt) -lB exp(lt) = 0

And the algebraic system:

(3 - l) A + 2B = 0

-5A + (1 - l) B = 0

And then to the form: (A - l) D = 0 =

(3 - l………2)

(-5………1 - l )

(-5………1 - l )

And then expand to obtain the
characteristic equation:

l

^{2}– 4l + 13 = 0
Using the quadratic formula one
can determine the roots of this system (with b = -4, a =1, and c = 13) are 2

__+__3i. Now, set l = 2 + 3i in the algebraic system to get:
(1 – 3i) A + 2B = 0

-5A
+ -(1 – 3i) B = 0

A simple, nontrivial solution of this system (using the
method to solve for A, B we employed before) yields: A = 2 and B = -1 + 3i.
Using these one obtains the complex solutions:

x = 2 exp (2 + 3i)t

y = (-1 + 3i) exp (2 +
3i)t

Which can alternatively be written (using Euler’s formula) :

x = exp (2t) [(2 cost 3t) + i(2 sin 3t)]

y = exp(2t)[(-cos 3t – 3 sin 3t) + i (3 cos 3t – sin 3t)]

Since both the real and imaginary parts of this solution are themselves solutions of the system, we also obtain the two real solutions:

x = 2 exp t cos 3t

y = - exp 2t (cos 3t
+ 3 sin 3t)

and also:

x = 2 exp t sin 3t

y = -exp 2t (3cos 3t
- sin 3t)

x = 2exp (2t) (c1 cost 3t + c2 sin 3t)

y = exp(2t)[c1(-cos 3t – 3 sin 3t) + c2 (3 cos 3t – sin 3t)]

Problems for Math Mavens:

1) The
other root of the characteristic equation was: 2 – 3i. Set l = 2
– 3i and find out what the general solution is to the system illustrated in the
post.

2) Find
the complex and general solutions of the system:

dx/dt = 5x – 2y

dy/dt = 4x - y

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