Solve the system Dy/dt=x,dx/dt=-4y

dx/dt = -4y
dt/dx = -1/(4y)
dy/dx = dy/dt(dt/dx) = x(-1/(4y)) = -x/(4y)
4y dy = -x dx
∫ 4y dy = ∫ -x dx
2y^2 = -x^2/2 + c
y^2 = -x^2/4 + c/2
y = (-x^2/4 + c/2)^(1/2)

To solve the system you have to obtain both y and x as functions of t.
Differentiating dy/dt=x with respect to t gives d^y/dt^2=dx/dt=-4y
so d^2y/dt^2+4y=0 which has general solution y=Acos(2t)+Bsin(2t)
so x=dy/dt=-2Asin(2t)+2Bcos(2t) where A and B are constants.

(dy/dt)(dt/dx) = dy/dx = x(1/(-4y)) = -x/(4y)
ie

4y dy = -x dx
2y^2 = -(1/2)x^2 + k
y^2 = -x^2 + 2k

or y^2 = C - x^2 where C is some constant

dy/dt = x
dx/dt = -4y

(dy/dt) / (dx/dt) = x / (-4y)
(dy/dx) = x / (-4y)
-4y * dy = x * dx
-2y^2 = (1/2) * x^2 + C
y^2 = (-1/4) * x^2 + C
y = sqrt(C - (1/4) * x^2)