00:01
Okay, so for a, we need to sketch the vector field f and then sketch some flow lines, and guess some what shape these flow lines appear to have? so the given vector field is f -obxy equals 1 and x.
00:28
So let's draw the vector field first, and then we'll put some flow lines over the vector field.
00:36
Y x and then the vector field for this is approximately something like this okay so let's draw the flow line so in order to draw the flow line we're just going to have to follow the vector field so that looks something like this and then something like that okay, let's move on to b.
01:58
Okay, first of all, i want to guess the shape of these flow lines.
02:05
So the shape, it looks sort of like a function, f -bx, equals c1 plus c2 times x squared.
02:23
Looks something like that.
02:28
Okay, now b, when the parametric equation is given by x of t and y of t, we need to find the differential equations that is satisfied by the vector field, the given vector field.
02:50
So by the definition of the flow lines, we need to have, the velocity vector which equals the derivative of the position vector i'm going to write the term out equals the vector field the given vector field.
03:25
Okay so now the given vector field is one and then x let's take the argument x of t...