Yes, that's exactly what you want. Since y'= f(y), with no "x", isoclines are horizontal lines, in this case, y= .
This in my opinion should be easy. I'm just not sure why it isn't.
The question reads,
Consider the initivial Value problem dy/dx = sin(y)
subject to Initial conditions y(X)=Y
Where X and Y are constants.
It asks us to determine the ODE's isoclines and sketch its direction field in the range x betwen [-2,2]
I normally breeze through IVP problems. But this one has me confused
normally you set y'=c a constant, to set against constant slopes
then re-arrange to leave yiso= a function
but in this case c=sin(y)
which can't be what we want is it?
It says drawing the isoclines corresponding to the slopes -1,0 and +1 should suffice to draw the direction field.
Any idea what I am looking at integral curves? sorry this question has me doing this
I presume you are looking at direction fields because the problem told you to!
In any case, if you are to draw "isoclines corresponding to the slopes -1,0 and +1", then you should draw (light) vertical lines at , , and . Note that will be more than 3 lines! For example, sin(y)= 0 for y= 0, , , etc. At a number of points on those lines, draw short horizontal (slope= 0) line segments. sin(y)= 1 at y= , , , etc. Draw several short line segments on those vertical lines with slope 1. Similarly, sin(y)= -1 for y= , , , etc. Draw short line segments on those vertical lines with slope -1. Those short line segments represent the "direction field".
(By the way, this really has nothing to do with an "initial value problem" because no "initial value" is given. And, anyway, the distinction between an "initial value problem" and a "boundary value problem" is only relevant for equation of higher order than 1.)