1) Find eigenvalues of matrix A={(2,3);(-1,-2)}
Characteristic polynomial looks as follows:
(2-λ)(-2-λ)+3=0
λ^{2}-4+3=0
λ^{2}=1^{
}
λ_{1}=-1
λ_{2}=1
2) Find eigenvectors of A;
a) For λ_{1}=-1;
Solve {(3,3);(-1,-1)}(a_{1},a_{2})=0
You will get 3a_{1}-a_{2}=0 or (a_{1},a_{2})=(1,3)
First eigenvector is (1,3), corresponding to λ_{1}=-1.
b) For λ_{2}=1;
Solve {(1,3);(-1,-3)}(a_{1},a_{2})=0
You will get a_{1}-a_{2}=0 or (a_{1},a_{2})=(1,1)
Second eigenvector is (1,1), corresponding to λ_{2}=1.
The solution of a homogenous system looks like:
x(t)=c_{1}(1,3)e^{-t}+c_{2}(1,1)e^{t}.
In order to solve inhomogenous system, we may now assume that c_{1} and c_{2} depend on t and plug this into the inhomogenous equation.
We will get:
c_{1}(t)'e^{-t}(1,3)+c_{2}(t)'e^{t}(1,1)-c_{1}(t)(1,3)e^{-t}+c_{2}(t)(1,1)e^{t}={(2,3);(-1,-2)}c_{1}(t)(1,3)e^{-t}+{(2,3);(-1,-2)}c_{2}(t)(1,1)e^{t}+(e^{t},t)=-(1,3)c_{1}(t)e^{-t}+(1,1)c_{2}e^{t}+(e^{t},t)
Some terms cancel and we obtain:
c_{1}(t)'e^{-t}(1,3)+c_{2}(t)'e^{t}(1,1)=(e^{t},t)
Solving for c_{1}(t) and c_{2}(t) yields:
c_{1}(t)=(t-1)e^{t}/2-t/2+C_{1};
c_{2}(t)=t(1+e^{-t}/2)+te^{-2t}/4+e^{-2t}/8+C_{2};
Thus the general solution is:
x(t)=C_{1}(1,3)e^{-t}+C_{2}(1,1)e^{t}+[(t-1)/2-te^{-t}/2](1,3)+[te^{t}+t/2+¼*(t+½)e^{-t}](1,1)