Please help me with this math problem?

Question

Express the general solution of x'=(1, 2, 3, 0, 1, 2, 0, -2, 1)x in terms of real-valued functions. (this is 3x3 matrix, 1, 2, 3 on the left, 0, 1, 2 in the middle, 0, -2, 1 on the right.) Please help me step by step.

Answer 1

Sun,

 

Kirill showed you that the general complex solution is of the form

 

x=C₁s₁e^k₁t+C₂s₂e^k₂t+C₃s₃e^k₃t

 

with

k₁=1, s₁=(2,-3,2)

k₂=1-2i, s₂=(0,1,i)

k₃=1+2i, s₃=(0,-1,i)

 

The first term,

 

C₁s₁e^k₁t=C₁(2,-3,2)e^t,

 

is already real and exactly of the form of Hassan's answer. So it suffices to show how the other two terms turn real.

 

We have

 

C₂s₂e^k₂t+C₃s₃e^k₃t=C₂(0,1,i)e^(1-2i)t+C₃(0,-1,i)e^(1+2i)t

 

Let's pull out an e^t :

=e^t (C₂(0,1,i)e^-2it+C₃(0,-1,i)e^2it)

 

We can ignore the first component, which is zero anyway for these two terms.

 

Let's look at the second (x₂) component: it is

=e^t(C₂e^-2it - C₃e^2it)

 

Now let's use Euler's identity (see Kirill's answer) on both exponentials and collect all sine and cosine terms, then the last expression becomes

=e^t( (C₂-C₃) cos(2t) + i(-C₂+C₃) sin(2t) )

 

Let us define

 

c₂:=C₂-C₃ and c₃:=i(-C₂+C₃)

 

then the second component becomes

=e^t( c₂ cos(2t) +c₃ sin(3t) ),

 

which is now of the real form Hassan gave.

 

The big C's were complex, the little c's are real.

 

I will leave it to you to do the same thing with the third component (x₃). With the same definition of the little c's, it will be of the form Hassan gave,

e^t( c₂ sin(2t) - c₃ cos(2t) ).

 

Finally, put x back in vector form.

Answer 2

Hello Sun,

 

For your reference, since the others haven't settled on an answer, your solution will be:

 

x = c₁( 2, -3, 2 )e^t + c₂e^t( 0, cos(2t), sin(2t) ) + c₃e^t( 0, sin(2t), -cos(2t) ).

 

Clearly both methods above/below will work.  Having briefly looked over the last few questions you have asked, my guess is you may want to stick to the more straightforward approach begun by Kirill.  Andre's method is a bit more "advanced" and I would suggest coming back to view it once you are on firm footing with the other one.

 

Regards,

Hassan H.

Answer 3

1) Find eigenvalues of the matrix A;

 

Form a matrix A-kI, where I is the unit matrix (1 on diagonal, 0 otherwise). In this case,

A-kI={(1-k,2,3);(0,1-k,2);(0,-2,1-k)}

Solve the equation

det|A-kI|=0;

Expand determinant along the first row, since there are two zeroes there. You will get:

 

(1-k)*((1-k)²+4)=0

 

It immediately follows that k-1=0 or k²-2k+5=0; The first equation gives k₁=1; The second one gives

k₂=1-2i; k₃=1+2i; where i=√(-1); 

 

2) Find eigenvectors for the matrix A; Eigenvectors satisfy the equation As=ks; where k is the eigenvalue. Therefore, we need to solve the system of equations

(A-kI)s=0; where 0 is the zero vector, for each eigenvalue.

 

For the first eigenvalue we will end up with the following system:

 

{(0,2,3);(0,0,2);(0,-2,0)}s=0; As a result, you have two equations: 2s₁-2s₃=0 and 3s1+2s₂=0. Here s₁ is the first component of the solution vector, s₂ is the second one, and so on.

We can choose arbitrarily, but it is convenient to have s₁=2; s₂=-3; s₃=2.

 

Repeating the same for the other two eigenvalues, we will get:

For the eigenvalue k₂=1-2i; s₂=(0,1,i); 

For the eigenvalue k₃=1+2i; s₃=(0,-1,i);

 

The general solution is in the form:

 

x=C₁s₁e^k₁t+C₂s₂e^k₂t+C₃s₃e^k₃t;

 

You can also use the fact that e^±it=cos(t)±i*sin(t); You will end up with the following solution:

 

x=C₁*(2,-3,2)e^t+e^t[C₂*(0,-1,0)sin(2t)+C₃*(0,0,1)cos(2t)]