# Slope Fields and Determing Behavior of Any Solution

• Bashyboy
In summary, the differential equation y' = y^2 has a solution curve that depends on the initial value of y at t = 0. If the solution curve passes through a point with y > 0, then as t approaches infinity, y will also approach infinity. However, if the solution curve passes through a point with y < 0, then as t approaches infinity, y will approach 0. It is also possible for the solution curve to pass through the point (0,0), in which case y will remain at 0 as t approaches infinity.

## Homework Statement

The differential equation is y' = y^2
Draw a direction field for the given differential equation.
Based on the direction field, determine the behavior of y as t →∞. If this behavior depends
on the initial value of y at t =0, describe this dependency.

## The Attempt at a Solution

Okay, so I plotted the slope field by evaluating the derivative at several different y-values. This is what I observed:

The solution curve certainly depends on the initial y-value when t=0. If particular solution y(t) to the DE has a solution of the form (t=0, y>0) (passes through a point), then as t----> infinity,
y(t)----> infinity. On the other hand, if some particular solution y(t) passes through a point of the form (t=0, y<0), then as t--->infinity, y(t)----> 0

Are these correct observations; and have I used terminology and notation correctly?

Also, when considering solutions that pass through the negative portion of the y-axis, will some approach y=0 more quickly than others?

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I am confused as to why you would say "On the other hand" for the initial value of y< 0 then say that its behavior is exactly the same as for y> 0.

And you seem to have left out an important special case- what if y= 0 for some t? And if y goes to infinity even if y is negative, how does it pass y= 0?

What you have pointed out, I believe I have fixed.

In the case that some solution passes through (0,0), then as t---> infinity, y will remain zero, right?