Systems of Differential Eq.

a) The system of diff. eqns:

dx/dt = 3x + 2y ... (1)

dy/dt = -2x - y ... (2)

Taking diff. to (1):

d2x/dt2 = 3dx/dt + 2dy/dt

= 3dx/dt + 2(-2x - y)

= 3dx/dt - 4x - 2y

= 3dx/dt - 4x - (dx/dt - 3x) by (1)

= 2dx/dt - x

d2x/dt2 - 2dx/dt + x = 0

Char. eqn: λ2 - 2λ + 1 = 0 with double roots λ = 1

So x = (At + B) et where A and B are constants.

2y = dx/dt - 3x

= (At + B)et + Aet - 3(At + B)et

= [(A - 2B) - 2At]et

y = [(A/2 - B) - At]et

b) The diff eqns are:

t dx/dt = 3x + 2y ... (1)

t dy/dt = -2x - y ... (2)

With t = es, we have dx/dt = e-s dx/ds and dy/dt = e-s dy/ds

So the eqns become:

dx/ds = 3x + 2y

dy/ds = -2x - y

By the result of (a):

x = (Cs + D) es and y = [(C/2 - D) - Cs]es where C and D are constants.

Finally:

x = (C ln t + D) t and y = [(C/2 - D) - C ln t] t

c i) Adding the equations in (I):

2t dx/dt = 6x + 4y + 2t2

t dx/dt = 3x + 2y + t2

Subtracting them:

2t dy/dt = -4x - 2y

t dy/dt = -2x - y

which comes to the results.

ii) Seperating them:

t d[u(t)]/dt = Au(t) gives x = (E ln t + F) t and y = [(E/2 - F) - E ln t] t

according to the result in (b), where E and F are constants.

t d[u(t)]/dt = (t2 0) gives:

t dx/dt = t2, dx/dt = t, i.e. x = t2/2 + G where G is a constant

t dy/dx = 0, y = H where H is a constant.

So the general solution is:

x = (E ln t + F) t + t2/2 + G

y = [(E/2 - F) - E ln t] t + H