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3 Answers

This is an inhomogeneous system, with the inhomogeneity being (et,t).
 
The general solution will be the sum of the general solution to the homogeneous system,
 
xh'=(2, 3, -1, -2) xh ,
 
and any particular solution to the inhomogeneous system, xp :
 
x = xh + xp
 
You solve the homogeneous system as usual by finding the eigenvalues and eigenvectors first. The eigenvalues of the matrix (2, 3, -1, -2) are -1 and 1, the corresponding eigenvectors are (1,3) and (1,1).
 
Therefore,
 
xh = c1 (1,3) e-t + c2 (1,1) et .
 
To find a particular solution, you should use the method of undetermined coefficients. Let me know if you are unfamiliar with this method. You assume that xp is of a form similar to your inhomogeneous term (et, t), namely
 
xp = A t + B et + C tet + D
 
Here A, B, C, and D are the undetermined coefficients that we determine by substituting this solution into the original equation. They are vectors with components A=(A1, A2), B=(B1,B2), C=(C1,C2), and D=(D1,D2). (Don't confuse the big C's with the little c's from the homogeneous solution.) Substitute all of this into your equation. Separate out the upper and lower equations for t, et, tet, and 1, and get 8 equations for the 8 unknowns A1 ... D2. After a lot of algebra you will find
 
A1 = 1, A2 = 2, B1  = -1/4, B2= -3/4, C1 = C2 = 3/2, D1 = 0, D2 = -1.
 
Therefore,
 
xp = (t - 1/4 et + 3/2 tet, 2t - 3/4 et + 3/2 tet - 1)
 
Add this to xh to get the final answer, x = xh + xp.
 
 
 
 
1) Find eigenvalues of matrix A={(2,3);(-1,-2)}
 
Characteristic polynomial looks as follows:
(2-λ)(-2-λ)+3=0 
λ2-4+3=0
λ2=1
λ1=-1
λ2=1
 
2) Find eigenvectors of A;
 
a) For λ1=-1; 
Solve {(3,3);(-1,-1)}(a1,a2)=0
You will get 3a1-a2=0 or (a1,a2)=(1,3)
First eigenvector is (1,3), corresponding to λ1=-1.
 
b) For λ2=1;
Solve {(1,3);(-1,-3)}(a1,a2)=0
You will get a1-a2=0 or (a1,a2)=(1,1)
Second eigenvector is (1,1), corresponding to λ2=1.
 
The solution of a homogenous system looks like:
 
x(t)=c1(1,3)e-t+c2(1,1)et.
 
In order to solve inhomogenous system, we may now assume that c1  and c2 depend on t and plug this into the inhomogenous equation.
 
We will get:
 
c1(t)'e-t(1,3)+c2(t)'et(1,1)-c1(t)(1,3)e-t+c2(t)(1,1)et={(2,3);(-1,-2)}c1(t)(1,3)e-t+{(2,3);(-1,-2)}c2(t)(1,1)et+(et,t)=-(1,3)c1(t)e-t+(1,1)c2et+(et,t)
 
Some terms cancel and we obtain:
c1(t)'e-t(1,3)+c2(t)'et(1,1)=(et,t)
 
Solving for c1(t) and c2(t) yields:
 
c1(t)=(t-1)et/2-t/2+C1;
c2(t)=t(1+e-t/2)+te-2t/4+e-2t/8+C2;
 
Thus the general solution is:
 
x(t)=C1(1,3)e-t+C2(1,1)et+[(t-1)/2-te-t/2](1,3)+[tet+t/2+¼*(t+½)e-t](1,1)