Solutions to Extra DE 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.   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.