# Direction field of $y' = y^2-4$

I have the ODE $$y' = y^2-4$$

I want to draw the direction field and draw the solutions that satisfy $$y(0)=-4$$ and $$y(0)=0$$ without solving the equation.

So i am writing $$y^2-4 = c$$ and then i start giving values to c in order to calculate y.

$$c = 0, y=_{-}^{+}2$$ $$c = 1, y=_{-}^{+}\sqrt{5}$$ $$\vdots$$

Then how am i drawing the direction field and the integral curves?

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