## Monday, July 15, 2013

### Solution to Non-Homogenous DE Problem

Find the general solution for the system:

dx/dt = 2x + y + 2t

dy/dt = 2x + 3y + t

Alert readers will see that the homogenous form for the above has already been solved (July 1st), e.g. http://brane-space.blogspot.com/2013/07/solution-to-problem-1-of-june-30-set.html

So we know the homogenous part of the soln is:

x = exp (4t) + exp (t)
and
y = 2 exp(4t) - exp (t)

Now, rewrite the forcing terms for the system as F(t) =
[2] t
[1]

And try a particular soln. of the system possessing the same form so that:   Xf = D1 t

Where D1 =
[a2]
[b2]

Then we proceed by setting:  X’f =  C (Xf) + C1 t

C is the original coefficient matrix, e.g.
[2.....1]
[2 ....3]

And C1 is the (2, 1) column matrix, with a factor t. Then we will need to solve (by undetermined coefficients):

[a2]
[b2]   = C [ D1 t]

Writing all this out one should obtain:
2a2 + b2 + 2 = 0
2a2 + 3b2 + 1 = 0

For which we obtain: a2 = -5/4  and b2 = ½

The forcing components of the solution are then:  Xf1 =  -5 t/4  and Xf2 = t/2

The general solution is then:

x = c1exp (4t) +  c2exp (t) – 5t/4

y = 2c1 exp(4t) -  c2exp (t) + t/2