Exact vs decimal input and imaginary solutions to ODEs

Questions
,
1

When I enter the following command I get imaginary solutions:

R = 5; **L = 0.05**; EMF = 5 Cos[120 t];

cur1 = DSolve[{R i [t] + L i'[t] == EMF, i[0] == 1}, i, t][[1, 1, 2]][ t]

Solution:
(0.409836 - 4.3368110^-17 I) E^(-100. t) ((1.44 + 2.5819610^-16 I) + (1. + 0. I) E^(100. t) Cos[120. t] + (1.2 + 1.26982*10^-16 I) E^(100. t) Sin[120. t])

But if I enter the following i get real solutions:

R = 5; **L = 1/20**; EMF = 5 Cos[120 t];
cur1 = DSolve[{R i [t] + L i'[t] == EMF, i[0] == 1}, i, t][[1, 1, 2]][ t]

Solution:

1/61 E^(-100 t) (36 + 25 E^(100 t) Cos[120 t] + 30 E^(100 t) Sin[120 t])

Any help understanding this would be appreciated. I am currently using Mathematica v11.1. But I have the same issue on Wolfram Cloud.

2

You can use Chop function: Chop—Wolfram Documentation

In[1]:= R=5;L=0.05;EMF=5 Cos[120 t];
In[3]:= sol=Chop@DSolve[{R i[t]+L i'[t]==EMF,i[0]==1},i,t]
Out[3]= {{i->Function[{t},0.409836 E^(-100. t) (1.44 +1. E^(100. t) Cos[120. t]+1.2 E^(100. t) Sin[120. t])]}}
In[4]:= i[2]/.sol[[1]]
Out[4]= 0.59849
3

A matrix exponential is being computed. With approximate numeric input it is using methods that show explicit numeric fuzz in the imaginary (oscillatory) part. As noted, Chop can be used to remove it.

4

Thanks Shenghui Yang and Daniel Lichtblau, for the solutions and explanation.

Ashish

5

The situation is not specific to ODE solving. see Help: tutorial/ExactAndApproximateResults. it effect most anything you do in Mathematica.