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For a system of differential equations:

$$ x'=x^2-1,\\ y'=-y $$

(1)Draw the solution curve that starts at the intial point $Y(0)=(x(0),y(0))$.

(2)For the above solution curve $Y(t)=(x(t), y(t))$, draw the $(t, y(t))$ graph.

I know the answer of the first one, it's a straight line from origin the the point $(-1,0)$, but I am not sure what's the second question is about.

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You do realize there is a difference between $(x(0),y(0))$ and $(0,0)$, right? How can you look at the slope field and know where on there $t=0$ occurs? – JohnD Dec 16 '12 at 20:34
I don't see the picutre – yiyi Dec 21 '12 at 0:04
up vote 0 down vote accepted

The picture is hard for me to see well, but based on what you say: if the solution $(x(t),y(t))$ looks like a horizontal line, what does that tell you about $(t,y(t))$?

You could also get the answer from observing that the system is decoupled and $y'=-y$. Once you've deduced what $y(0)=y_0$ is, the graph of $y(t)$ is determined.

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I know that y(0)=0. Just one more question, so the graph of t,y(t) looks like y=-0.5x^2 with x greater than zero! YEah! But we ignored x(t) simply because of the initial condition or we can just ignore it regardless the initial condition? – user48601 Dec 16 '12 at 20:41
$y$ is a function of $t$, not $x$. $x$ is a (separate) dependent variable in this system. Rather $y'=−y\implies y(t)=Ce^{−t}$. Then $y(0)=0\implies C=0$ so $y(t)=0$. – JohnD Dec 16 '12 at 20:57

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