8.2 Const-Coeff Homog Systems: 
X'[t]=A.X[t]

Soln of X'[t]=A.X[t]  n linearly independent evecs.

(Subscript[X, i])[t]=(Subscript[K, i]) (e^((Subscript[λ, i]) t))  where A.(Subscript[K, i])=(Subscript[λ, i])(Subscript[K, i])  eigenstuff

Gen sol (∑_(i=1)^n)(Subscript[C, i]) (Subscript[X, i])[t] IF there are n linearly independent evecs.

Soln of X'[t]=A.X[t] missing eigenvector

If there is a double eigenvalue λ and only one eigenvector K then (Subscript[X, 1])[t]=K (e^(λ t)) is a soln.
You get a second soln by plugging 
	(Subscript[X, 2])[t]=K t (e^(λ t))+P (e^(λ t))
into the Eq 
	(Subscript[X, 2]) '[t] =K λ t (e^(λ t))+K  (e^(λ t)) +P λ (e^(λ t))=A.K t (e^(λ t))+K  (e^(λ t)) +P λ (e^(λ t))	and
	A.(Subscript[X, 2])[t]=A.K t (e^(λ t))+A.P (e^(λ t))
So (Subscript[X, 2]) is a soln if K  (e^(λ t)) +P λ (e^(λ t))=A.P (e^(λ t)) or 
	(A-λ Id) P=K.
The process can be continued if there are more missing eigenvectors.

Soln of X'[t]=A.X[t]  Complex Eigenvalues and Eigenvectors

If A is real then the eigenvalues are either real or come in complex conjugate pairs
	λ=α±β i
The eigenvectors come the same way
	K=(Subscript[B, 1])± (Subscript[B, 2])i
The two solutions 
	(e^((α + β i)t))((Subscript[B, 1])+(Subscript[B, 2]) i) and (e^((α - β i)t))((Subscript[B, 1])-(Subscript[B, 2]) i)
work but they have i=(Sqrt[-1]) in them. This is OK in EE but not in many other fields!

Soln of X'[t]=A.X[t]  Real Solns from Complex Evals

The trick is that (e^((α + β i)t))=(e^(α t))(Cos[β t]+i Sin[β t]). Sooooooooooooooo
	({{(e^((α + β i)t))((Subscript[B, 1])+(Subscript[B, 2]) i), =, (e^(α t))(Cos[β t]+i Sin[β t])((Subscript[B, 1])+(Subscript[B, 2])i)                                                }, {(e^((α + β i)t))((Subscript[B, 1])+(Subscript[B, 2]) i), =, (e^(α t))(((Subscript[B, 1])Cos[β t]-(Subscript[B, 2])Sin[β t])+i ((Subscript[B, 2])Cos[β t]+(Subscript[B, 1])Sin[β t]))}})
Doing the same thing on the other one
	({{(e^((α -β i)t))((Subscript[B, 1])-(Subscript[B, 2]) i), =, (e^(α t))(((Subscript[B, 1])Cos[β t]-(Subscript[B, 2])Sin[β t])-i ((Subscript[B, 2])Cos[β t]+(Subscript[B, 1])Sin[β t]))}})
If you add/subtract these two solutions you get two real solutions
	({{(Subscript[X, 1])[t], =, (e^(α t))((Subscript[B, 1])Cos[β t]-(Subscript[B, 2])Sin[β t])}, {(Subscript[X, 2])[t], =, (e^(α t))((Subscript[B, 2])Cos[β t]+(Subscript[B, 1])Sin[β t])}})

Why (e^(α + β i))=(e^α)(e^(β i))=(e^α)(Cos[β]+i Sin[β])

From Calc II

TableForm[{Series[(E^x),{x,0, 6}],
Series[Cos[x],{x,0,6}],
Series[Sin[x],{x,0,6}]}]

Usefull Commands

Mathematica can analytically solve all the systems of equations we are going to learn how to solve! Lets look at a 3×3.

Clear[x1,x2,x3,x4,x5,X]
A=(({{1, 1, 0}, {0, 1, 1}, {0, 0, 2}}));
X[t_]:={x1[t],x2[t],x3[t]}
MatrixForm[
X[t]/.DSolve[{
X'[t]==A.X[t]
},
X[t],
t][[1]]
]

Now with IC

Clear[x1,x2,x3,x4,x5,X]
A=(({{1, 1, 0}, {0, 1, 1}, {0, 0, 2}}));
X[t_]:={x1[t],x2[t],x3[t]}
MatrixForm[
X[t]/.DSolve[{
X'[t]==A.X[t],
X[0]=={1,2,3}
},
X[t],
t][[1]]
]

Semi-Manually

Eigensystem[A]

The two eigenvectors give two solutions!  I am going to write them down.

MatrixForm[C[3] (E^(2t)){1,1,1}+C[1](E^(1 t)){1,0,0}]

We find the third by solving (A-λ Id)P=K where {λ,K} are the messed up eigen value/vector pair so we solve (A-1 Id)P={1,0,0} for P and build the last bit t (e^(λ t))K+(e^(λ t))P

LinearSolve[A-1 IdentityMatrix[3],{1,0,0}]

General solution is

MatrixForm[C[3] (E^(2t)){1,1,1}+C[1](E^(1 t)){1,0,0}+C[2](t (E^(1 t)){1,0,0} + (E^t){0,1,0})]

If we have initial conditions (say X[0]={1,2,3}) we plug in t=0 and get

(({{1C[1]+0 C[2]+1C[3]}, {0C[1]+1 C[2]+1 C[3]}, {0C[1]+0C[2]+1 C[3]}}))=(({{1, 0, 1}, {0 , 1, 1}, {0, 0, 1}})).(({{C[1]}, {C[2]}, {C[3]}}))=(({{1}, {2}, {3}}))

LinearSolve[(({{1, 0, 1}, {0 , 1, 1}, {0, 0, 1}})),{1,2,3}]

Pugging these back into the general solution gives the particular soln satisfying the IC

(({{(e^t) C[1]+(e^t) t C[2]+(e^(2 t)) C[3]}, {(e^t) C[2]+(e^(2 t)) C[3]}, {(e^(2 t)) C[3]}}))⟶(({{(e^t) (-2)+(e^t) t (-1)+(e^(2 t)) (3)}, {(e^t) (-1)+(e^(2 t)) (3)}, {(e^(2 t)) (3)}}))