I was confronted with the following problem on my mid-term paper and I've got no idea how to solve this, I tried using the eigenvalues method, but it ultimately failed . Can anyone, please, give a complete solution to this ? I really want to se a proper solution so I can understand better the reasoning at hand..
$y' = 5y +4z $
$z'=-4y -3z$
Hint:
Substitute the first equation in the second
$$ z' = -4y -\frac{3}{4}(y'-5y)$$
Yet, from the first equation
$$z' = \frac{1}{4}(y'' - 5y')$$.
Hint:
Add both equations:
$$y'+z'=y+z\quad\Longrightarrow\quad y+z=ce^t$$
Now substitute.
this problem is more difficult than i thought because the coefficient matrix has a repeating root. i need a little set up first. let $x$ be a two dimensional column vector and $A$ a $2 \times 2$ real matrix. consider the linear differential equation $$\frac{dx}{dt} = Ax$$
first we establish that:
(a) if $\lambda$ is an eigenvalue and $x_0 \neq 0$ is a corresponding eigenvector pf $A,$ then $x = x_0e^{\lambda t}$ is a solution.
(b) if $x_1 \neq 0$ is such that $(A - \lambda)^2x_1 = 0, (A-\lambda)x_1 = x_0 \neq 0,$ then $x = x_0 te^{\lambda t} + x_1 e^t$ is a solution. (a way to see this is to assume that $A$ has eigenvalues $\lambda \pm \epsilon.$ by (a) $x_0(e^{\lambda t + \epsilon t} - e^{\lambda t - \epsilon t}) /{ 2 \epsilon} = x_0te^{\lambda t}$)
in your case $$ A= \pmatrix{5 & 4\\-4 & -3}$$ the eigenvalues of $A$ are given by $$det(A- \lambda I) = det \pmatrix{5 - \lambda & 4\\-4 & -3 - \lambda} = (\lambda - 1)^2$$ we will pick $$x_1 = \pmatrix{1 \\ 0}, x_0 = \pmatrix{4 \\ -4}$$
the two linearly independent solutions of $\frac{dx}{dt} = Ax$ is $$\left\{ \pmatrix{1 \\-1}e^t, \pmatrix{4t+1\\-4t}e^t \right\} $$
