Solution to (2) (Part B)

We left off with:

[2 ……..-4] [A]

[1……..-2] [B]

 

Whence: 2 A - 4B = 0 and A – 2B = 0

 
So: A =2 and B = 1, and  c1 =

[2]
[1]

 

Now we take:

[2 ……..-4] [A2]

[1……..-2] [ B2]

So that: 2A2 – 4 B2 = 2  and A2 – 2B2 = 1

Then: A2 = 1 and B2 = 0 to satisfy the equation. And c2=

[A2]
[B2]

=

[1]

[0]

 

The solutions are then:

X1 = c1 exp (t)

X2 = c2 exp(t) + c1 t exp (t)

We would like to write these in terms of solutions in x1 and x2, so we need to pick out the A, B column vectors that correspond to the x1, x2 solutions bearing in mind the general forms:

Then: x1 =   2 exp (t) is a first solution

while a second solution is:  x1= exp (t) + 2 t  exp (t))

Also:

x2 = exp (t) is a first solution

while a second solution is:  x2=   t exp (t)

Check the 2^nd DE of the system:   dx2/dt = x1 - x2

by using the two first solutions:  x1 =   2 exp (t)    x2 =   exp (t)

Then:  dx2/dt =  d/dt (exp (t)) = exp (t)

And:  x1 - x2  =  2 exp (t)   - exp (t) = exp (t)

Now repeat for:

dx1/dt = 3x1 - 4x2

Using the same initial solutions.

dx1/dt =  2 exp (t)  [Recall: d/dx (au) = a (du/dx)]

where a is the constant (2) and u is the function exp (x) (in our case exp (t))

and: 3x1 - 4x2 = 3 [2 exp(t)] - 4 exp (t) 

= 6 exp(t) - 4(exp(t)) = 2 exp(t)

Extra Problems:

1) By the principle of superposition you can add the solutions, such that:

X = c1 X1  + c2 X2

Use that to write out the additional solutions in x1 and x2 (do NOT confuse with X1, X2!).


2) Check to see the solutions you obtained in (1) satisfy the same system of equations as I showed in the example preceding the Problems (for the initial pair of solutions).