Non-Homogenous Linear Systems of Differential Equations

In the case of non-homogenous linear systems we also include an additional “forcing term”  for each equation, so the system will look like this:

dx/dt = 6x + y + 6t (on -¥, ¥)

dy/dt = 4x + 3y – 10t + 4

And we refer to the terms beyond those in x, y on the right as the “forcing terms”. The standard procedure entails first solving the homogenous system, then attending to the forcing terms using the method of undetermined coefficients. Proceeding, we find that the eigenvalues are determined from: (A - lI)D = 0 =

[(6 - l)  …..1]

[4 ……..(3-  l)]

=  l² – 9l  + 14   or:  (l -2) (l - 7) = 0

So the roots are 2 and 7. Insert l1  = 2:

[4.....1] [A]
[4  ...1 ] [B]

Leading to 4A + B = 0 so A = 1, B = -4 and K1=

[1]
[-4]

Do the similar procedure for the root l1  = 7 and one finds:K2 =

[1]
[1]

The so-called “complementary function” is then expressed:

X1 = K1 exp (2t)  and X2 = K2 (exp 7t) or:

x= c1 exp (2t) + c2 exp (7t)

y = -4c1 exp (2t) + c2 exp (7t)

Now, rewrite the forcing terms as F(t) =

[6] t
[-10]

+

[0]
[4]

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

Where D1 =

[a2]
[b2]   and D =

[a1]
[b1]

Then we proceed by setting:  X’f = C (Xf) + C1 t  +  C2 where C is the original coefficient matrix, e.g.

[6.....1]
[4 ...-3] 

And C1 is the (6, -10) column matrix (see above) while C2 is the (0, 4) column matrix. Then we will need to solve (by undetermined coefficients):

[a2]
[b2]   = C [ D1 t + D] + C1t + C2

Writing all this out one should obtain:

(6a2 + b2 + 6)t + 6a1 + b1 – a2 = 0

And:

(4a2 + 3b2 – 10)t + 4a1 + 3b1 – b2 + 4 = 0

Which essentially is just messy algebra to configure and solve for a1, b1, a2, b2. First, solve the pair of equations in t:

6a2 + b2 + 6 = 0

4a2 + 3b2 – 10 = 0

The correct result is a2 = -2 and b2 = 6 which interested readers can check. Then solve the pair of equations for:

6a1 + b1 – a2 = 0

4a1 + 3b1 – b2 + 4 = 0

Insert values known for a2, b2:

6a1 + b1  + 2  = 0

4a1 + 3b1 -2 = 0

Solving for a1, b1 gives: a1 = - 4/7 and b1 = 10/7

We obtain from this the solution including the forcing:  Xf1 = -2 t  - 4/7  and Xf2 =  6t + 10/7

The general solution is then:

x= c1 exp(2t) + c2 exp (7t) – 2t – 4/7

y = -4c1 exp (2t) + c2 exp (7t) + 6t + 10/7

Problem for the Math Maven:

Find the general solution for the system:

dx/dt = 2x + y + 2t

dy/dt = 2x + 3y + t