1. ## IVP, slightly confused

This in my opinion should be easy. I'm just not sure why it isn't.

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)

therefore yiso=sin-1(c)

which can't be what we want is it?

2. Yes, that's exactly what you want. Since y'= f(y), with no "x", isoclines are horizontal lines, in this case, y= $\displaystyle sin^{-1}(c)$.

3. Originally Posted by HallsofIvy
Yes, that's exactly what you want. Since y'= f(y), with no "x", isoclines are horizontal lines, in this case, y= $\displaystyle sin^{-1}(c)$.
Hmmm never had a question like this, so how would i go about marking the direction field?

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

4. o and thanks

5. 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 $\displaystyle sin^{-1}(-1)$, $\displaystyle sin^{-1}(0)$, and $\displaystyle sin^{-1}(1)$. Note that will be more than 3 lines! For example, sin(y)= 0 for y= 0, $\displaystyle \pi$ $\displaystyle -\pi$, $\displaystyle 2\pi$, etc. At a number of points on those lines, draw short horizontal (slope= 0) line segments. sin(y)= 1 at y= $\displaystyle \pi/2$, $\displaystyle -3\pi/2$, $\displaystyle 5\pi/2$, etc. Draw several short line segments on those vertical lines with slope 1. Similarly, sin(y)= -1 for y= $\displaystyle -\pi/2$, $\displaystyle 3\pi/2$, $\displaystyle 7\pi/2$, 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.)

6. So my diagram should have isoclines that are horizontal defining teh slopes of

y=

And then also light vertical lines at the points where the slopes equals -1,0 and +1?

but surely a vertical line

sin(y)= 0 for y= 0, ,

at that point will go through all of them. what with them being y co-ordinates?