#1 dy/dx = (y-1)²(2+x) goes through (-1,2) I need to find the equation of the curve and the equations of the asymptotes to it.

Group the variables so that dy is with the y-terms, and dx is with the x-terms:

dy/(y-1)^2 = (2+x)dx

[(y-1)^(-2)]dy = (2+x)dx

Integrate bote sides,

-1(y-1)^(-1) = 2x +(x^2)/2 +C

1/(1-y) = 2x +(x^2)/2 +C ----------(i)

at (-1,2), or when x = -1, y=2, so,

1/(1-2) = 2(-1) +((-1)^2)/2 +C

-1 = -3/2 +C

C = 1/2

So, into (i),

1/(1-y) = 2x +(x^2)/2 +1/2

Multiply both sides by 2,

2/(1-y) = 4x +x^2 +1

(1-y) = 2/(4x +x^2 +1)

y = 1 -2/(x^2 +4x +1)

y = [x^2 +4x +1 -2]/(x^2 +4x +1)

y = (x^2 +4x -1)/(x^2 +4x +1) -------the equation of the curve, answer.

Vertical asymptotes:

x^2 +4x +1 = 0

x = {-4 +,-sqrt[4^2 -4(1)(1)]} /(2*1)

x = -3.732 or -0.268 ------------------two vertical asymptotes.

Horizontal asymptotes:

Lim(x-->infinity)[(x^2 +4x -1)/(x^2 +4x +1)]

= Lim(x-->infinity)[(1 +4/x -1/(x^2)) / (1 +4/x +1/(x^2))]

= 1

So, y = 1 is a horizontal asymptote

Oblique asymptote: None, because the numerator (x^2 +4x -1) is not one degree higher than the denominator (x^2 +4x +1).