# Question:work around for a problem in dsolve, numeric?

## Question:work around for a problem in dsolve, numeric?

Maple
Of course, there are big differences in dsolve and dsolve numeric. But also there is a difference that is subtle. It would be nice if this subtle one were not there. It may be that there is an easy work-around. Of course, the complication that is at the root of the problem may not be subtle at all. Look at the results of these two situations. First situation. > restart; > h[0]:=t->(1+t)*t^2*cos(t): > sol1:=dsolve({diff(y(t),t)=-h[0](t-1),y(0)=h[0](0)},y(t)): > h[1]:=unapply(rhs(sol1),t): > sol2:=dsolve({diff(y(t),t)=-h[1](t-1),y(1)=h[1](1)},y(t)): > h[2]:=unapply(rhs(sol2),t): > plot([[t,h[0](t),t=-1..0],[t,h[1](t),t=0..1], [t,h[2](t),t=1..2]],color=blue); Second situation > restart; > k[0]:=t->(1+t)*t^2*cos(t): > sol1:=dsolve({diff(y(t),t)=-k[0](t-1),y(0)=k[0](0)},y(t), numeric, output=listprocedure): > k[1]:=subs(sol1,y(t)): > plot([[t,k[0](t),t=-1..0],[t,k[1](t),t=0..1]],color=blue); > sol2:=dsolve({diff(y(t),t)=-k[1](t-1),y(1)=k[1](1)},y(t), numeric, output=listprocedure): > k[2]:=subs(sol2,y(t)): > plot([[t,k[0](t),t=-1..0],[t,k[1](t),t=0..1], [t,k[2](t),t=1..2]],color=blue); Of course, where I want to make this delay model is where the ode's are not linear and cannot be solved analytically.
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