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Find the general solution?

Find the general solution of x'=(3, 2, -2, -2)x. (This is a matrix, 3 and 2 on the left, -2 and -2 on the right.)
 
Answer: x=c1(1, 2)e^-t+c2(2, 1)e^2t
 
After c1 and c2 are the matrixes, 1 on the top, 2 at the bottom, and 2 at the top, 1 at the bottom. Please show your work step by step.
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2 Answers

Write the system as two coupled equations:
x'=3x-2y
y'=2x-2y
Solve the first equation for 2y:
2y=3x-x'
and substitute this into the second equation:
y'=2x-3x+x'=x'-x
Now take the derivative of the first equation:
x''=3x'-2y'
and substitute what we just found for y':
x''=3x'-2(x'-x)=x'+2x, or
x''-x'-2x=0.
This equation you know how to solve with a characteristic polynomial:
r²-r-2=(r-2)(r+1)=0,
so r=-1 and 2, and the solution for x is
x=c1e-t+c2e2t , with
x'=-c1e-t+2c2e2t
 
To find y, substitute this x into the equation 2y=3x-x' and simplify:
 
2y=3(c1e-t+c2e2t )-(-c1e-t+2c2e2t )=4c1e-t+c2e2t , or
y=2c1e-t+½c2e2t
 
 
1) Find eigenvalues for the matrix A=((3,-2),(2,-2)). Here and later first inner parentheses denote first row, second--second row. In order to do it, one need to solve secular equation det|A-k*I|=0, where k is the eigenvalue to be found and I is unit matrix, I=((1,0),(0,1)). In our case we will obtain:
 
det|((3-k,-2),(2,-2-k))|=0. From this we will get:
(3-k)(-2-k)-2*(-2)=0 or (k-3)*(k+2)+4=0 or k2-k-2=0
Solutions of this quadratic equation are given by: 
k1=1/2*[1+√(1-4*(-2))]=(1+√9)/2=2
k2=1/2*[1-√(1-4*(-2))]=(1-√9)/2=-1
 
2) Now, let us find eigenvectors corresponding to each of the eigenvalues. We plug in first eigenvalue, k=2, into the matrix equation above to obtain:
K1=A-k1*I=A-2I=((1,-2),(2,-4)); Now we need to solve K1(a1,a2)=0 for vector (a1,a2). If we use the explicit form of matrix K1, we will obtain two identical equations:
a1-2a2=0 and 2a1-4a2=0. Those have infinitely many solutions, corresponding to infinitely many vectors, which all point in the same direction, but differ in length. Let us pick the most obvious solution, a1=2; a2=1; Thus, first eigenvector is (2,1). Often eigenvectors are normalized, then this eigenvector would be (2,1)/√5.
Next, we do the same for the other eigenvalue, k2. Upon substitution and solving the equation, we will obtain the second eigenvector, (1,2).
 
3) General solution to a differential equation in question is in the form:
C1*(first eigenvector)*Exp[k1t]+C2*(second eigenvector)*Exp[k2t]
Here k1 shall correspond to its eigenvector and so shall k2! Do not mix them up!
 
In our case, (2,1) eigenvector corresponds to k1=2 and (1,2) eigenvector corresponds to k2=-1, thus we obtain:
x(t)=C1*(2,1)*e2t+C2*(1,2)*e-t;