Question

(a) sketch the slope field for the differential equation,and (b) use the slope field to sketch the solution that passes through the given point.
Y'=-x-2 (-1,1)

Answer 1

I tried plugging -1 in, but i get -1 not 1. I also tried finding y by integrating and it still doesn't work

Answer 2

I know the answer, but i dont get how to do it

Answer 3

http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/lecture-1-the-geometrical-view-of-y-f-x-y/

Answer 4

brb, i will watch it

Answer 5

take your time, its a very good video

Answer 6

I get how to do them when there is also a y in the equation, then you just plug in points

Answer 7

\[y'=-x-2\]

first derivative gives the slope of tangent line at a particular point on the curve, yes ?

Answer 8

yes, then you use that slope to draw the slope field

Answer 9

so would the slope at (-1,1) be -1?

Answer 10

supppose the slope is \(c\), then the isocline is
\[c = -x-2\]

which is same as
\[x=-(c+2)\]

Answer 11

which gives a family of vertical lines

Answer 12

Okay, i understand so far

Answer 13

fix slope, \(c = 0\) and you get
\[x = -(0+2) \implies x = -2 \]

so you draw small line segments with slope=\(0\) along the vertical line \(x=-2\)

Answer 14

fix different values for \(c\) and draw as many slope segments as you can

Answer 15

So instead of plugging in different points, you plug in different slope?

Answer 16

how would we find the y coordinate though?