In[]:=
∂
t,t
z[t]==-z[t]
3/2
(
2
z[t]
+
2
(1/2(1+eSin[2πt]))
)
Out[]=
′′
z
[t]-
z[t]
3/2

1
4
2
(1+eSin[2πt])
+
2
z[t]

In[]:=
Clear[z]
In[]:=
NDSolveValue
∂
t,t
z[t]==-z[t]
3/2
(
2
z[t]
+
2
(1/2(1+1/2Sin[2πt]))
)
,z'[0]==0,z[0]==.1,z,{t,0,50}
Out[]=
InterpolatingFunction
Domain: {{0.,50.}}
Output: scalar

In[]:=
Plot[%[t],{t,0,50}]
Out[]=
In[]:=
NDSolveValue
∂
t,t
z[t]==-z[t]
3/2
(
2
z[t]
+
2
(1/2(1+1/2Sin[2πt]))
)
,z'[0]==0,z[0]==110,z,{t,0,50},WorkingPrecision->50,MaxSteps->10^6
NDSolveValue
:Error test failure at t == 14.498434611246519702738998021067586674474612928340​; unable to continue.
Out[]=
InterpolatingFunction
Domain: {{0,14.498434611246519702738998021067586674474612928340}}
Output: scalar

In[]:=
Plot[%271[t],{t,0,14}]
Out[]=
In[]:=
Plot[%268[t]-%271[t],{t,0,14}]
Out[]=
In[]:=
NDSolveValue
∂
t,t
z[t]==-z[t]
3/2
(
2
z[t]
+
2
(1/2(1+1/2Sin[2πt]))
)
,z'[0]==0,z[0]==110,z,{t,0,50},WorkingPrecision->50,MaxSteps->10^7
NDSolveValue
:Error test failure at t == 14.498434611246519702738998021067586674474612928340​; unable to continue.
Out[]=
InterpolatingFunction
Domain: {{0,14.498434611246519702738998021067586674474612928340}}
Output: scalar
Data not saved. Save now

In[]:=
NDSolveValue
∂
t,t
z[t]==-z[t]
3/2
(
2
z[t]
+
2
(1/2(1+1/2Sin[2πt]))
)
,z'[0]==0,z[0]==110,z,{t,0,50},WorkingPrecision->100,MaxSteps->10^7
NDSolveValue
:Maximum number of 10000000 steps reached at the point t == 33.18123334822323633740312871192426536903543612743393882219782850457165604095999089884086411934051054.
Out[]=
InterpolatingFunction
Domain: {{0,33.18123334822323633740312871192426536903543612743393882219782850457165604095999089884086411934051054}}
Output: scalar
Data not saved. Save now

In[]:=
Plot[%268[t]-%276[t],{t,0,33}]

Eccentricity 1

NDSolveValue
∂
t,t
z[t]==-z[t]
3/2
(
2
z[t]
+
2
(1/2(1+Sin[2πt]))
)
,z'[0]==0,z[0]==.1,z,{t,0,50}
Out[]=
InterpolatingFunction
Domain: {{0.,50.}}
Output: scalar

In[]:=
Plot[%[t],{t,0,50}]
Out[]=

Programmability

For what value of z[0] is the trajectory easy to predict?
In[]:=
NestList[x|->FractionalPart[2x],FromDigits[{{1,0,1,1,0,1,1,1,0},-1},2],20]
Out[]=

183
512
,
183
256
,
55
128
,
55
64
,
23
32
,
7
16
,
7
8
,
3
4
,
1
2
,0,0,0,0,0,0,0,0,0,0,0,0
In[]:=
IntegerPart[2%]
Out[]=
{0,1,0,1,1,0,1,1,1,0,0,0,0,0,0,0,0,0,0,0,0}