- #1
ajayguhan
- 153
- 1
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)...
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)...
