[tex]\dfrac{\mathrm dy}{\mathrm dt}=-4x\implies x=-\dfrac14\dfrac{\mathrm dy}{\mathrm dt}\implies\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}[/tex]
Substituting this into the other ODE gives
[tex]-\dfrac14\dfrac{\mathrm d^2y}{\mathrm dt^2}=-4y\implies y''-16y=0[/tex]
Since [tex]x(t)=-\dfrac14y'(t)[/tex], it follows that [tex]x(0)=-\dfrac14y'(0)=4\implies y'(0)=-16[/tex]. The ODE in [tex]y[/tex] has characteristic equation
[tex]r^2-16=0[/tex]
with roots [tex]r=\pm4[/tex], admitting the characteristic solution
[tex]y_c=C_1e^{4t}+C_2e^{-4t}[/tex]
From the initial conditions we get
[tex]y(0)=5\implies 5=C_1+C_2[/tex]
[tex]y'(0)=16\implies-16=4C_1-4C_2[/tex]
[tex]\implies C_1=\dfrac12,C_2=\dfrac92[/tex]
So we have
[tex]\boxed{y(t)=\dfrac12e^{4t}+\dfrac92e^{-4t}}[/tex]
Take the derivative and multiply it by -1/4 to get the solution for [tex]x(t)[/tex]:
[tex]-\dfrac14y'(t)=\boxed{x(t)=-\dfrac12e^{4t}+\dfrac92e^{-4t}}[/tex]
