# 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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## 2 Answers

You are given the equation for y' which tells you the slope of y at any point. So try picking various values of y to see what direction y should be moving. For example at t = 1, say y = 3, then y' = 5. So at the point {1, 3} you can draw a sharp upward pointing vector (arrow on vector would point to about 1 o'clock). Then repeat this for various points on your y and t plot until you have enough vectors to give you an idea of the solution plot.

If you have access to mathematica, it can do this for you :)

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What riotburn said. Also, to draw the solution curves, start at the point (0, -4) and trace the path that follows the arrows. Do the same for (0, 0).

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