00:03
We're given the differential equation, and we're asked to find a slope field of this differential equation, and to find three solutions.
00:15
Differential equation is y prime equals x minus y plus one.
00:23
Now we see that if we have some slope c in the real numbers, so that y prime is equal to c, and it follows it c is one equal to x.
00:46
Minus y plus 1 or that's y, it'll be equal to x plus 1 minus c.
01:30
This is the equation of a line with slope 1 in y intercept 1 minus c.
01:40
So i'm graphing the slope field, you see in particular, if we take c to be 1, there we have a slope of 1, then the line y equals x, all of the slopes on that line will be 1, so if you follow the line y equals x, then we'll get slip of 1 at 0, 0, slip of 1 again at 1.
03:17
1, 1, slip of 1 again at 2, 2, and the other points in between on that line, and so on.
03:59
We have that c is say equal to 0, then we have the line, y equals x plus 1, all the points and that is going to have slope of 0.
04:13
So we have a slope of 0 at 0, 0, 1, and at 1, 2, 2, 3, and so on.
04:38
We can fill in slopes and line 20s at 0.
05:01
If we take c to be 2, say, then we get the line y equals x plus negative 1.
05:17
It's going to have y intercept as negative 1.
05:20
This will have a slope of 2, so even steeper.
06:28
If c is equal to 3, or if she's equal to negative 1, let's say we'll get the line y equals x plus 2.
06:41
Now we'll have the downward slope at 02, negative 1.
06:49
We can continue this along the line.
06:54
We take c to be negative 2.
07:45
So even steeper, we get the line y equals x plus 3.
07:53
So a very steep line at 0 ,3.
08:12
You know along that line.
08:14
We have these steep downward slopes.
08:29
I think it's kind of easy to see the pattern here so that as we get further and further away from the line, y equals x plus 1, our slopes get steeper and steeper.
08:49
But they're almost vertical...