Question

Let y = f(x) be a particular solution to the differential equation dy/dx = 1 + y^2 - 2x - 2xy^2 with f(2) = 0
Write an equation for the line tangent to the graph of y=f(x) at x=2

Answer 1

@sourwing ? :3

Answer 2

random guess-> y = -2(x-2) ? :P

Answer 3

did you get
dy/dx = (1+y^2) (1 - 2x) ?

Answer 4

yes...
for the "Find the particular solution y = f(x) to the differential equation with the initial condtion f(2) = 0. (Hint: Factor out by grouping)" part

Answer 5

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

arctan(y) = x - x^2 + C

Answer 6

oh wait, actually you don't have to solve the differential equation
dy/dx is a slope in terms of x and y.
Given (x,y) = (2,0)
then dy/dx = (1 + 0^2) (1 - 2(2)) = -3

using point slope form
y - 0 = -3(x-2)

Answer 7

oh right okay :) thanks

Answer 8

np

Answer 9

it's a multipart problem tho
how would I do the "Find the particular solution y = f(x) to the differential equation with the initial condtion f(2) = 0. (Hint: Factor out by grouping)" part?

Answer 10

arctan(y) = x - x^2 + C, and (x,y) = (2,0). Plug the point in and solve for C

Answer 11

alright :) thanks again ^_^

Answer 12

C = 2?

Answer 13

oh yeah it's 2 :D

Answer 14

lol ok xD
so I would write it as arctany = -x^2 + x + 2 ?
or try to isolate y somehow?

Answer 15

if you want to

Answer 16

it looks like the question wants you to solve for y since it said y = f(x)