Current[t_]:= (0.08*Sin[4*t]-2*Cos[4*t]+2*Exp[-0.08*t])/16
Plot[Current[t],{t,0,40}]

We wish to study this by means of the Euler method, which is a numerical method.
The idea is that the next value  at T + delT can be approximated by evaluating the formul

(d/dt)Current = -0.08*Current + Sin[4*T]

and letting

T= (i-1)*delT



Curr[i] := Curr[i-1] + (-0.08*Curr[i-1] + Sin[4*(i-1)*delT])delT

Curr=0
delT:=0.03
For[i=1, i<800, i++, Curr=Curr+(-0.08*Curr+Sin[4*(i-1)*delT])*delT;Cu[i]=Curr]

Cu[5]

For j = 800, time is

Tihm:= 0.03*800
Tihm
DiscretePlot[Cu[j],{j,800}]

Force the above plot to go to 40 by changing the code.

change delT to 0.06 and comapre the solutions

Then solve the problem  for

Curr2=0
delT2:=0.0
For[i=1, i<200, i++, Curr2=Curr2+(-0.08*Curr2+Sin[4*(i-1)*delT2])*delT2;Cu2[i]=Curr2]
DiscretePlot[Cu2[j],{j,200}]
DiscretePlot[Cu[j],{j,400}]

Solve the problem 


I’ + 0.2I = Sin[3*t]

by adapting the above code.