1. Part (a)
Because they ask for two equal steps starting at -1 and ending at 0 we know that x is .5
They tell us to start at x= -1
Now we just plug in values according to the formula at the top of the table to complete the Euler’s Method Approximation
And that f(x)=y=2 so our first y coordinate is 2
(x,y)
dy/dx x
y=(dy/dx)( x)
(x+ x, y+ y) 4 .5
(-.5,4)
(-1,2) 2
(-.5,4) .5
.25
(0,4.25)
Answer!

2. Part (b) P2(x)= 2 + 4(x+1) + P2(x)= 2+4(x+1)-6(x+1)2
To find the second-degree Taylor polynomial about x=(-1) we need to find the values of f(-1), f ’(-1), f ’’(-1)
y0=2
y’=6(-1)2 – (-1)2(2)
y’=4
y’’= -12
(-12)(x+1)2 2!
P2(x)= (x+1) +
P2(x)= 2+4(x+1)-6(x+1)2

3. ∫ Part (c) dy/dx = 6x2-x2y -ln|6-y| = (1/3)x3+C dy/dx = x2(6-y)
-y=-6+4 e-1/3x3-1/3
-ln(4) = (-1/3)+C
= (x2)dx dy
(6-y)
y=6- 4 e-1/3x3-1/3
C= (1/3)-ln(4)
ln|6-y| = -(1/3)x3-(1/3)+ln(4)
= (x2)dx dy
(6-y) ∫
|6-y|=(e-1/3x3-1/3) (4)
-ln|6-y| = (1/3)x3+C
6-y=+4 e-1/3x3-1/3