_{1}(1, 2)e^-t+c

_{2}(2, 1)e^2t

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=c_{1}(1, 2)e^-t+c_{2}(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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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=c_{1}e^{-t}+c_{2}e^{2t} , with

x'=-c_{1}e^{-t}+2c_{2}e^{2t }

To find y, substitute this x into the equation 2y=3x-x' and simplify:

2y=3(c_{1}e^{-t}+c_{2}e^{2t} )-(-c_{1}e^{-t}+2c_{2}e^{2t} )=4c_{1}e^{-t}+c_{2}e^{2t
, or}

y=2c_{1}e^{-t}+½c_{2}e^{2t}

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 k^{2}-k-2=0

Solutions of this quadratic equation are given by:

k_{1}=1/2*[1+√(1-4*(-2))]=(1+√9)/2=2

k_{2}=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:

K_{1}=A-k_{1}*I=A-2I=((1,-2),(2,-4)); Now we need to solve K_{1}(a_{1},a_{2})=0 for vector (a_{1},a_{2}). If we use the explicit form of matrix K_{1}, we will obtain two identical equations:

a_{1}-2a_{2}=0 and 2a_{1}-4a_{2}=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, a_{1}=2; a_{2}=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, k_{2}. 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:

C_{1}*(first eigenvector)*Exp[k_{1}t]+C_{2}*(second eigenvector)*Exp[k_{2}t]

Here k_{1} shall correspond to its eigenvector and so shall k_{2}! Do not mix them up!

In our case, (2,1) eigenvector corresponds to k_{1}=2 and (1,2) eigenvector corresponds to k_{2}=-1, thus we obtain:

x(t)=C_{1}*(2,1)*e^{2t}+C_{2}*(1,2)*e^{-t};

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