Need help with the following:
$$\begin{align*}\dfrac{\mathrm{d}x}{\mathrm{d}t} &= −5x − y \\ \dfrac{\mathrm{d}y}{\mathrm{d}t} &= 4x − y \end{align*}$$
With initial conditions: $x(1) = 0, y(1) = 1$
I converted it into
$$\begin{align*} (D+5)x+y &= 0\\ -4x+(D+1)y &= 0 \end{align*}$$
And got the solutions $x(t) = c_1e^{-3t} + c_2te^{-3t}$ and $y(t) = c_3e^{-3t}+c_4te^{-3t}$.
I tried to substitute into the first equation and solve for $c_3, c_4$ in terms of $c_1, c_2$ but I must have done something wrong as I am not getting the correct answer. Any help is appreciated.
By elimination we should have
$$x'=-5x-y\implies y=-x'-5x \qquad y'=-x''-5x'$$ $$-x''-5x'=4x+x'+5x \implies x''+6x'+9x=0\implies x(t) = c_1e^{-3t} + c_2te^{-3t}$$
and then from $y=-x'-5x$ we have
$$y(t) = 3c_1e^{-3t} +3c_2te^{-3t}-c_2e^{-3t}-5c_1e^{-3t} -5c_2te^{-3t}=(-2c_1-c_2)e^{-3t} -2c_2te^{-3t}$$
thus we can find $c_1$ and $c_2$ by initial conditions.