# Direction field of y'=cos(πx)

1. Dec 13, 2013

### ajayguhan

This isn't a home work, so only I'm posting here

When we solve y'=cos(πx) we'll get y=sin(πx)/π +C. But plotting direction field seems little confusing for me

Since y' is a function of cosine , the value of y' must be within [-1,1].

If i took y' as 0 and tried to plot it and I'm getting set of points such as (.....-2.5, -1.5, -0.5, 0.5, 1.5, 2.5,....)

I can't draw a isocline since here it is discontinuois.

Similarly by setting y' has 1, -1 we'll get points such as (......-4, -2, 0, 2, 4.....) & (......-5, -3, -1, 1, 3, 5....)

Now how the solution curve should be plotted from the direction field, without isocline (i mean here isocline is just set of points not a curve).......

2. Dec 15, 2013

### ajayguhan

No one knows where I'm wrong...?