Determining vector in a direction field

[kpou]

Homework Statement

Determine the vector in the direction field( in the form of < 1, ? >) correspond-
ing to the given point.
y' = 1 + 2ty at (2, 0), at (3, -2).

Homework Equations

The Attempt at a Solution

y'-2ty=1
p(t)=-2t ; u(t)=e^(-t^2)
e^(-t^2)-2te^(-t^2)=e^(-t^2)
e^(-t^2)=integral(e^(-t^2)dt)+c

I am confused where to go from here. I think I am supposed to set up the integral definitely. When i do from 0->0 I get c=2.
From here I am not sure where to go. Do I set up the integral from 0->1 (since I am trying to find a vector with t =1)? When I do this it comes out to <1, 2.63> What made me doubt this was how I am supposed to set up the problem at point (3,-2).

Do I plug in t=0 and for the integral do 0->1 ?

And plug in t=3 and do 3->4 integral ?

This does not really seem right though -.-

My weakness may lie in the fact that I have no notes on how to find a vector <1, ?>.

 

[HallsofIvy]

you are asked to find the direction field. That means that you do NOT want to solve the equation and do NOT want to integrate. All the information you need is in the differential equation.

At (2, 0) (which I am going to interpret as t= 2, y= 0) dy/dt= 1+ 2(2)(0)= 1 so your vector must have slope 1: the two components must be equal, if the vector is <x, y>, then y/x= 1.

At (3, -2), dy/dt= 1+ 2(3)(-2)= -11. The slope must be -11: if the vector is <x, y>, then y/x= -11.

In fact, this whole problem is pretty close to being trivial!