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. Solution: We note from the blog post (July 3rd) that x1 = 2 exp (t) is a first solution (for x1) while a second solution is:
x1= exp (t) + 2 t exp (t)
Then we add to obtain the superposed solution:
x1= 2 exp(t) + exp (t) + 2 t exp (t) = 3 exp(t) + 2t exp (t)
Similarly, for x2: x2 = exp (t) is a first solution, while a second solution is: x2= t exp (t)
Then the superposition of solns is: x2 = exp(t) + t exp (t)
2) Check to see the solutions you obtained in (1) satisfy the system of equations:
a) dx1/dt = 3x1 - 4x2
b) dx2/dt = x1 - x2
In the first case (a):
dx1/dt = d/dt [3 exp (t) + 2t exp(t)] =
3 exp(t) + 2t exp(t) + 2 exp(t)
= 5 exp (t) + 2t exp (t)
And: 3x1 = 3 [3 exp (t) + 2t exp(t)] = 9 exp (t) + 6 t exp (t)
4x2 = 4 [exp(t) + t exp(t)] = 4 exp (t) + 4t exp (t)
So: 3x1 - 4x2 =
[9 exp (t) + 6 t exp (t)] - [4 exp (t) + 4t exp (t)]
= 5 exp(t) + 2t exp (t)
So the solution satisfies the D.E.
In the second case (b):
dx2/dt = d/ dt [exp (t) + t exp (t)] =
exp (t) + exp (t) + t exp (t) = 2 exp (t) + t exp (t)
Meanwhile:
(x1 - x2) =
[3 exp (t) + 2t exp (t)] - [exp (t) + t exp (t)]
= [3 exp(t) - exp(t)] + [2t exp(t) - t exp(t)] = 2 exp(t) + t exp (t)
Which solution satisfies the D.E.
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