Find the general solution?

Question

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.

Answer 1

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+c₂e^2t , with

x'=-c₁e^-t+2c₂e^2t

 

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

 

2y=3(c₁e^-t+c₂e^2t )-(-c₁e^-t+2c₂e^2t )=4c₁e^-t+c₂e^{2t _{, or}}

y=2c₁e^-t+½c₂e^2t

 

 

Answer 2

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)*e^2t+C₂*(1,2)*e^-t;