# Question:How to plot a proc in maple 16

## Question:How to plot a proc in maple 16

Maple 16

I have a differential that I'm solving using rk4 on maple and I want to plot the resaulting points in an interval. How do I do that?

here is what I've already done:

> f := {y(0) = 0, diff(y(x), x) = cos(5*y(x))-x};

> p := dsolve(f, y(x), numeric, method = classical[rk4], stepsize = .1);
proc(x_classical)  ...  end;

> for i from 0 by .1 to 3 do p(i) end do;
[x = 0., y(x) = 0.]
[x = 0.1, y(x) = HFloat(0.09134174797303919)]
[x = 0.2, y(x) = HFloat(0.1566319772919618)]
[x = 0.3, y(x) = HFloat(0.194581375068523)]
[x = 0.4, y(x) = HFloat(0.21164724456423595)]
[x = 0.5, y(x) = HFloat(0.21461563189591115)]
[x = 0.6, y(x) = HFloat(0.2084355484472073)]
[x = 0.7, y(x) = HFloat(0.19628552691639553)]
[x = 0.8, y(x) = HFloat(0.18006011418673554)]
[x = 0.9, y(x) = HFloat(0.16078626173516788)]
[x = 1.0, y(x) = HFloat(0.1388990025222126)]
[x = 1.1, y(x) = HFloat(0.11439234722678368)]
[x = 1.2, y(x) = HFloat(0.08686281440786428)]
[x = 1.3, y(x) = HFloat(0.05543927201293955)]
[x = 1.4, y(x) = HFloat(0.018544097249920552)]
[x = 1.5, y(x) = HFloat(-0.026681001192892698)]
[x = 1.6, y(x) = HFloat(-0.08574884829506454)]
[x = 1.7, y(x) = HFloat(-0.17028751480382714)]
[x = 1.8, y(x) = HFloat(-0.3061861464355401)]
[x = 1.9, y(x) = HFloat(-0.5351701804890165)]
[x = 2.0, y(x) = HFloat(-0.8187863602434449)]
[x = 2.1, y(x) = HFloat(-1.0288715875633263)]
[x = 2.2, y(x) = HFloat(-1.1730821561000684)]
[x = 2.3, y(x) = HFloat(-1.3002163707859586)]
[x = 2.4, y(x) = HFloat(-1.4535355524105638)]
[x = 2.5, y(x) = HFloat(-1.694911173717997)]
[x = 2.6, y(x) = HFloat(-2.036983303280925)]
[x = 2.7, y(x) = HFloat(-2.309185573036668)]
[x = 2.8, y(x) = HFloat(-2.50089099925866)]
[x = 2.9, y(x) = HFloat(-2.69768426675822)]
[x = 3.0, y(x) = HFloat(-2.995102028138579)]

So I need to plot those points. Thanks in advance ﻿