00:01
We're given the differential equation, and the point in the plane, we ask to find the scope field of this equation and to draw the graph of a solution which contains this point.
00:14
The differential equation is y prime equals x plus y squared, and the point in the plane is the origin, zero, zero.
00:23
Finding the slope field of this equation, suppose c is an element of real numbers and that y prime is equal to c.
00:35
That c is equal to x plus y squared so that x is equal to negative y squared plus c.
00:57
You can think of these as horizontal parabolas that lie on the x axis facing to negative infinity and which have an x intercept.
01:23
Of 0c.
01:28
And we have the all points on this parabola, we'll have slope c.
01:35
So, start by drawing my x and y axes.
01:47
I want to emphasize the negative side of the x -axis, since that's what most of the parabola will be.
02:24
So in particular, if c is equal to 0, then we have x z -equal to negative y squared.
02:33
So we have a parabola to the point at 0 0 and also has a point at negative 1, 1, and negative 1, negative 1.
02:52
And all along this parabola, the points have slope 0.
03:21
Equal takes c to be, say, 1, then we'll have x equals negative y squared plus 1, and we have a point 1, 0.
03:40
On this parabola, which is a slope of 1.
03:47
We also have the point 0 ,1, and 0, and the and all along this parabola points have a slope of 1.
04:10
This parabola lies just outside our first parabola.
04:38
Then you see that the c gets larger and larger.
04:42
Our slopes are going to get steeper and steeper.
04:44
And so we get these shells of parabolas between steeply and steeper slopes on.
05:30
And the slink is still always positive, although it becomes nearly vertical, further and further, remove towards a bay.
06:20
Now, if c is somewhere between 0 and 1, we'll get out of parabola, but its slope will be between 0 and 1...