       Re: Finding branches where general solution is possible

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• Subject: [mg131864] Re: Finding branches where general solution is possible
• From: "Dr. Wolfgang Hintze" <weh at snafu.de>
• Date: Sat, 19 Oct 2013 04:12:16 -0400 (EDT)
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• References: <l3ljoe\$jt5\$1@smc.vnet.net> <l3no2j\$ov6\$1@smc.vnet.net>

```Am Donnerstag, 17. Oktober 2013 06:12:03 UTC+2 schrieb Dr. Wolfgang Hintze:
> 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?
>
> >
>
> >
>
> >
>
> > Regards
>
> >
>
> > 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) - E^(2*C)*Cos[2*#1])*
>
>                    Sec[#1/2]^4]]*
>
>
>
>               Sqrt[(-(2 - 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]) - #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) - E^(2*C)*Cos[2*#1])*
>
>                    Sec[#1/2]^4]]*
>
>
>
>               Sqrt[(-(2 - 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]) + #1/2) & ][
>
>     t + C]}}
>
>
>
> Best regards,
>
> Wolfgang

In order to make things (which Mathematica "knows") more tranparent consider that the differential equation is of the form

s''(t) = f(s(t)) g(s'(t))

Here variables can be separated as follows.
Letting s' = u and using s as the indepent variable instead of t we have

du/dt = du/ds ds/dt = du/ds u

and the differential equation becomes

du/ds u = f(s) g(u)

or

du u/g(u) = ds f(s)

This can be integrated once and gives (possibly implicitly) u as a function h of s and the constant C.

But this is another differential equation

ds/dt = h(s,C)

which can be integrated to give

Int(1/h(s,C)) = t+C

This is then the general solution (two constants of integration) in the inverted form t = t(s).

Best regards,
Wolfgang

```

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