Find the general solution of the firstorder system of ODEs x

Find the general solution of the first-order system of ODEs x\' = -x + 4y. y\' = -4x - y.

Solution

Write the equation in matrix-vector form, i.e. let v = (x, y) and v\' = (x\', y\') and then we have

v\' = Av

where A =

[-1, 4]
[-4, -1].

Then we find the Eigen-values with by finding the determinant of A minus m times the identity, and letting it equal 0: det(A - mI) =

| -1 - m, 4 |
| -4, -1 - m |

which is

(-1 - m)² - (-16) = 0

(m + 1)² + 16 = 0

(m + 1)² = -16

m + 1 = (+-)4i

m = -1 (+-)4i.

Now you need the eigenvectors: we solve (A - mI)v = 0. For the e-value -1 + 4i, we solve

(-1 - (-1 + 4i))v1 + 4v2 = 0
-4v1 + (-1 - (-1 + 4i))v2 = 0

4iv1 + 4v2 = 0 [1]
-4v1 + 4iv2 = 0 [2]

Using [2] gives v1 = iv2. So let v2 = s and v1 = is. Then the eigenvector for -1 + 4i is [is, s] = s*[i, 1] or just [i, 1].

Doing the same process with the eigen-value -1 - 4i, we get

(-1 - (-1 - 4i))v1 + 4v2 = 0
-4v1 + (-1 - (-1 - 4i))v2 = 0

giving

-4iv1 + 4v2 = 0 [1]
-4v1 - 4iv2 = 0 [2]

Using [1] shows that v2 = iv1, so let v1 = s and v2 = 1s. The second eigen-value becomes [1, i].

So the general solution of this is given by multiplying each eigenvector by a constant of integration and its solution given by the eigenvalue; i.e.

v = c1*[i, 1]*e^((-1 + 4i)t) + c2*[1, i]*e^((- 1 - 4i)t).

For the \'cos and sine\' form, use e^(it) = cos(t) + i*sin(t), i.e

v = e^(-t)*{c1*[i, 1]*e^(4it) + c2*[1, i]*e^(-4it)}

= e^(-t)*{c1*[i, 1]*(cos(4i) + i*sin(4i)) + c2*[1, i]*(cos(4i) - i*sin(4i))}

= e^(-t)*{[i*c1 + c2, c1 + i*c2]*cos(4t) + i*[i*c1 + c2, c1 + i*c2]*sin(4t)}

= e^(-t)*{[i*c1 + c2, c1 + i*c2]*cos(4t) + i*[i*c1 + c2, c1 + i*c2]*sin(4t)}

= e^(-t)*{[i*c1 + c2, c1 + i*c2]*cos(4t) + [-c1 + i*c2, i*c1 - c2]*sin(4t)}.

Or we can separate out to real and imaginary, so

= e^(-t)*{([c2, c1] + i*[c1, c2])*cos(4t) + (-[c1, c2]+ i* [c2, c1])*sin(4t)}

= e^(-t)*{[c2*cos(4t) - c1sin(4t), c1cos(4t) - c2sin(4t)] + i*[c1*cos(4t) + c2*sin(4t), c2*cos(4t) + c1*sin(4t)]}.