Question

using rulers method, step size of 0.5, compute approximate y-values for y-sub1, y-sub2, y-sub-3, and y-sub4 of the solution of the initial-value problem y-prime=y-2x, y(1)=0

Answer 1

i don't know where the x-knot and y-knot are coming from

Answer 2

well, \(x_0=1,y_0=0\) from your initial condition \(y_0=y(x_0)\) in this case \(y(1)=0\)

Answer 3

then you find \(x_1=x_0+h\) where \(h=1/2\), and \(x_2=x_1+h\), etc.

Answer 4

so x-knot=1 is from the y(1)?

Answer 5

yes precisely... P.S. it's 'x-naught' not 'x-knot'

Answer 6

so x-naught is a given and we find y-naught from the 0 of y(1)=0?

Answer 7

\(y_0\) is just \(y(x_0)\) so yes \(0\) in this case

Answer 8

so both naughts are from y(x-naught)=y-naught?

Answer 9

indeed

Answer 10

so, for example:$$y_0=0\\y_1=0+1/2~(0-2(1))=-1\\y_2=-1+1/2~(-1-2(1+1/2))=-3\\y_3=-3+1/2~(-3-2(1+1/2+1/2))=-13/2\\y_4=-13/2+1/2~(-13/2-2(1+1/2+1/2+1/2))=-49/4$$

Answer 11

http://www.mathscoop.com/image-gallery/slope-fields/y-2x5-55-5ffffff3300661-1120400ff0000100_53190.png