Ordinary Differential Equation

Question

The coefficient matrix A = [−1 +2; −2 +3] of the linear system

(1) x'(t) = −x(t) +2y(t)

(2) y'(t) = −2x(t) +3y(t)

has one real eigenvalue λ = 1 of multiplicity two, but the eigenspace E_λ = R⋅(1,1)^T is one-dimensional.

How do I find the general solution of (1)&(2)?

Answer 1

For a repeated eigenvalue the second solution is

(x₂, y₂)^T = (a, b)^T t e^t + (c, d)^T e^t.

If you plug in the new solution (this is not necessary it is okay to just use the result) then (a,b)^T = (1,1)^T.

(x₂, y₂)^T = (1, 1)^T t e^t + (c, d)^T e^t.

After plugging it in you also get, (A - I) (c, d)^T = (1,1)^T.

There is a general formula for this (A - λ₁ I) (a, b)^T = u₁, where A u₁ = λ₁ u₁.

This gives (a, b)^T as ((-2, 2)(-2, 2)) (c, d)^T = (1,1)^T.

This does not have a unique solution but don't stress out, the combination of this and the first solution will work together to give a unique solution to the initial conditions. I am going to choose (c, d)^T as (0,1/2)^T.

Then (x₂, y₂)^T = (1, 1)^T t e^t + (0, 1/2)^T e^t.

This is a little tricky to check so

Here is a check of the top equation,

https://www.wolframalpha.com/input?i=+%28d%2Fdt+t*exp%28t%29%29-%28-exp%28t%29*t%2B+2*%28exp%28t%29*t%2B1%2F2*exp%28t%29%29%29

and here is a check of the bottom

https://www.wolframalpha.com/input?i=%28d%2Fdt+t*exp%28t%29%2B1%2F2*exp%28t%29%29-%28-2exp%28t%29*t%2B+3*%28exp%28t%29*t%2B1%2F2*exp%28t%29%29%29

For fun, (x₂, y₂)^T = (1, 1)^T t e^t + (-1/2, 0)^T e^t, should also work, and it does.

Check of top equation.

https://www.wolframalpha.com/input?i=%28d%2Fdt+t*exp%28t%29%2B1%2F2*exp%28t%29%29-%28-2exp%28t%29*t%2B+3*%28exp%28t%29*t%2B1%2F2*exp%28t%29%29%29

Check of bottom equation.

https://www.wolframalpha.com/input?i=%28d%2Fdt+t*exp%28t%29%29-%28-2%28exp%28t%29*t-1%2F2*exp%28t%29%29%2B+3*%28exp%28t%29*t%29%29

Answer 2

You have a linear system with the coefficient matrix A:

A = [-1 2; -2 3]

The system is described by:

x'(t) = -x(t) + 2y(t) y'(t) = -2x(t) + 3y(t)

You've identified that there's one real eigenvalue, λ = 1, with a one-dimensional eigenspace Eλ.

This eigenspace corresponds to the eigenvector (1, 1)T.

To find the general solution of this system:

1. Use the eigenvector (1, 1)T associated with the eigenvalue λ = 1.
2. Find a linearly independent solution. Since you have a repeated eigenvalue, you'll need to find a generalized eigenvector. Let's call it V, and it's of the form [a; b].
3. Calculate the solution corresponding to this generalized eigenvector: e^(At) * V.
4. Combine these solutions to form the general solution:

General Solution: X(t) = c1 * e^(t) * [1; 1] + c2 * e^(t) * [a; b]

Here, c1 and c2 are arbitrary constants, and 'a' and 'b' depend on the specific values in your matrix A.

This is the general solution for your linear system. It incorporates both the eigenspace solution and the linearly independent solution associated with the repeated eigenvalue.

Best regards,

Benjamin M.