#Instructions: use this function to compute approximate numerical values of 
#solutions to initial value problems using the euler method
#
#if you want to solve dy/dt=f(t,y) for y(t0)=y0 with step size h,
#try eulerapprox(f,t0,y0,h,n) where n is the number of values you want to see
#
#for example, if y'=3+t-y and y(0)=1, and step size h=0.1, try
#eulerapprox(3+t-y,0,1,0.1,4);
#
eulerapprox:=proc(f,t0,y0,h,n) local yprev, ycurr, tprev, tcurr, i:
ycurr:=y0:
tcurr:=t0:

for i from 1 to n do
yprev:=ycurr:
tprev:=tcurr:

#If y_n=yprev and t_n=tprev, then...

#y_(n+1) = y_n + f(t_n,y_n)*h (equation 8 on page 103)
ycurr:=evalf(yprev+subs({y=yprev,t=tprev},f)*h): 

#t_(n+1)=t_n+h
tcurr:=tprev+h: 

print("when t=", tcurr, ", y is approximately =", ycurr):
od:

end: