# Math Help - [SOLVED] Slope Field

1. ## [SOLVED] Slope Field

Sketch the direction field of the differential equation

y′ = 2 − 3y + $y^2$

in the domain $x \epsilon$[0, 3], $y \epsilon$[−3, 3].

Using the direction field decide whether a solution y(x) tends to a finite limit when $x\rightarrow\infty$.

How does this limit depend on the value of the initial condition y(0) ?
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Im not sure how to get the values to draw this directional field. Do I just sub in y=-3.-2.-1.0.1.2.3 ? and why do I have the domain of x when there is no x in the equation?

2. Originally Posted by ronaldo_07
Sketch the direction field of the differential equation

y′ = 2 − 3y + $y^2$

in the domain $x \epsilon$[0, 3], $y \epsilon$[−3, 3].

Using the direction field decide whether a solution y(x) tends to a finite limit when $x\rightarrow\infty$.

How does this limit depend on the value of the initial condition y(0) ?
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Im not sure how to get the values to draw this directional field. Do I just sub in y=-3.-2.-1.0.1.2.3 ? and why do I have the domain of x when there is no x in the equation?
Someone else already asked the same problem, but said (s)he could sketch the direction field so I'll answer to you anyway.

At the point $(x,y)$, the slope is $2-3y+y^2$. It doesn't depend on $x$: it means that the slopes will be the same on any "column". The direction field is invariant by translation along the $x$-axis. So you have to sketch one column of slopes (for instance at $x=0$, or anywhere else), and copy it for a few other values of $x$.

How to draw the directions? The main point is to see where the slope is positive or negative, or zero. Thus, as a start, you should study the sign of $2-3y+y^2$. Then draw the horizontal slopes (those equal to 0). You can draw the maximum/minimum slope if any, and add a few slopes to fill your sketch in and get a better feeling of what the field looks like.