Compute the general solution for the given system of differential equations.

Question

Compute the general solution for the given system of differential equations.

$$\frac{d}{dt}X=\begin{bmatrix}-1 &-4\\ 2& 3\end{bmatrix}X$$

Answer 1

Set $$A=\begin{bmatrix} -1 & -4\\ 2 & 3\\ \end{bmatrix} $$ Compute $Z=L^{-1}({(sI-A)^{-1}})$ in which $L$ denotes the Laplace transform ans $I$ is the identity matrix($2*2$). If your initial value is $x_{0}$ your answer will be: $$x(t)=Z*x_{0}$$

Answer 2

The problem

$$ \frac{dX}{dt} = A X $$

Can be solved in the following way:

Be v1 and v2 the eigenvectors of A:

$$ A v_i = \lambda_i v_i $$

And be U the matrix whose columns are v1 and v2. Then,

$$ A U = \Lambda U $$

where $ \Lambda $ is the matrix with the respective eigenvalues in the diagonal.

If $ U^{-1} $ is the inverse of $ U $, whe can define $ X' = UX $, and write the original equation as

$$ \frac{dU^{-1} X}{dt} = U^{-1} A X = U^{-1} A U U^{-1} X $$

or

$$ \frac{d X'}{dt} = U^{-1} A U X' $$

But $ U^{-1} A U = \Lambda $, such that we have

$$ \frac{d X'}{dt} = \Lambda X' $$

And the problem can be solved in a decoupled way:

$$ X' = exp\left(\Lambda t\right) $$

Finally:

$$ X = U^{-1} exp\left(\Lambda t\right) $$