       Re: Finding branches where general solution is possible

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• Subject: [mg131883] Re: Finding branches where general solution is possible
• From: Narasimham <mathma18 at gmail.com>
• Date: Wed, 23 Oct 2013 23:45:14 -0400 (EDT)
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```> Am Mittwoch, 16. Oktober 2013 10:46:06 UTC+2 schrieb
> Narasimham:
> > For following function with period 2 Pi in which
> branches is it possible to get a general solution?
> > Narasimham
> >
> > DSolve[{si''[th] Tan[si[th]]==(1+si'[th])
> (1+2si'[th]),si==Pi/4 },si,th]
> >
> > NDSolve[{si''[th] Tan[si[th]]==(1+si'[th])
> (1+2si'[th]),si'==0,si==Pi/4},si,{th,0,6Pi}];
> >
> > SI[u_]=si[u]/.First[%];Plot[SI[th],{th,0,6Pi}]
> > DSolve::bvnul: For some branches of the general
> solution, the given boundary conditions lead to an
> empty solution.>>
>
> Version 8 of Mathematica has no difficulty in
> DSolve-ing the equation generally.
> The solution is given in terms of InverseFunction as
> follows:
>
> In:= eq =
>  Derivative[s][t] ==
> Cot[s[t]]*(1 + Derivative[s][t])*(1 +
> + 2*Derivative[s][t])
>
> Out= Derivative[s][t] ==
> Cot[s[t]]*(1 + Derivative[s][t])*(1 +
> + 2*Derivative[s][t])
>
> In:= DSolve[eq, s[t], t]
>
> Out= {
>    {s[t] ->
>    InverseFunction[
>      2*(-((I*E^C*Cos[#1/2]^3*
>               Log[2*I*E^C*Cos[#1] -
>                 Sqrt*Cos[#1/2]^2*
> Sqrt[(-2 + E^(2*C) -
> -2 + E^(2*C) - E^(2*C)*Cos[2*#1])*
>                    Sec[#1/2]^4]]*
>
> Sqrt[(-(2 - E^(2*C) +
> - E^(2*C) + E^(2*C)*Cos[2*#1]))*
>                 Sec[#1/2]^4]*Sin[#1/2])/
>             Sqrt[(-E^(2*C))*(2 - E^(2*C) +
> E^(2*C)*Cos[2*#1])*Sin[#1]^2]) -
> ])*Sin[#1]^2]) - #1/2) & ][
>     t + C]},
>    {s[t] ->
>    InverseFunction[-2*(-((I*E^C*Cos[#1/2]^3*
>               Log[2*I*E^C*Cos[#1] -
>                 Sqrt*Cos[#1/2]^2*
> Sqrt[(-2 + E^(2*C) -
> -2 + E^(2*C) - E^(2*C)*Cos[2*#1])*
>                    Sec[#1/2]^4]]*
>
> Sqrt[(-(2 - E^(2*C) +
> - E^(2*C) + E^(2*C)*Cos[2*#1]))*
>                 Sec[#1/2]^4]*Sin[#1/2])/
>             Sqrt[(-E^(2*C))*(2 - E^(2*C) +
> E^(2*C)*Cos[2*#1])*Sin[#1]^2]) +
> ])*Sin[#1]^2]) + #1/2) & ][
>     t + C]}}
>
> Best regards,
> Wolfgang

Thanks Wolfgang, I am having difficulty in simplifying this Inverse periodic function for a given boundary condition at t = 0 for #1. E.g., can I get a closed form/general expression for s as a function of t, at least for boundary condition s == Pi/2 ? ... so as to be able to plot the numerical case given from that expression ?

Regards
Narasimham

```

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