Math 2300 Section 005 – Calculus II – Fall 2010
Notes on Slope Fields – October 25–26, 2010
Slope Field Examples

Find an infinite collection of solutions to the differential equation
Solution: It's nearly impossible to ask for even a single solution to this differential equation without knowing more about it. You can graph out the slope field yourself (perhaps not quite as neatly as a computer or calculator, but you can do it). You get a slope field like this.
Now it appears a bit more obvious that we're dealing with circles centered at the origin. Let's see if that is actually correct by taking a generic circle, centered at the origin, and determining if it satisfies the differential equation. In what follows, we use implicit differentiation to view as a function of .
So, each circle centered at the origin satisfies the given differential equation.

The differential equation is graphed in the following slope field along with a particular solution (the solid line). An initial condition representing the given solution is . Find the equation representing the given solution.
Solution: We know that has the solution . But, here's where knowing some algebra can help us. Consider changing to
where is some constant. Then we also have , and so we have made this solution a bit more general. But, now, consider
for another constant . This also satisfies . At first, if you properly understand graph transformations, this seems to contradict the notion of a slope field. (We can shift left and right using the constant , and we can ALSO expand/shrink vertically using the constant .) It seems like these two constants could conflict and allow us to produce a function not fitting in our slope field. Let's assume for now. Then you can find some such that
So, this is really equivalent to just selecting the appropriately. Likewise, if then we write and use the same reasoning to write
Thus, if we have some nonzero coordinate in an initial condition, we need to only consider a single constant. Since we know that our initial condition has a positive coordinate, we must then find a constant such that
Taking the natural log of both sides tells us that and so . Thus, the equation satisfying this differential equation plus our initial condition is

Use the following slope field for the differential equation to approximate the value of .
Solution: Since is a solution to the differential equation , we may use the initial condition from to follow a path through the slope field until we are above .
It appears as though the curve representing will reach a vertical height of around 0.65 to 0.7 when . As it turns out, the precise value of about .