I have the equations

x[t] == Exp[m t] && (2 t + 1) x''[t] + 2 (2 t - 1) x'[t] - 8 x[t] == 0

Please tell me which function to use to solve it and find the parameter m from the first equation that will solve the second,

I used DSolve on the second equation and found m = -2, but maybe there is another way.

  • $\begingroup$ Simplify[DSolve[{(2 t+1)x''[t]+2(2 t-1)x'[t]-8 x[t]==0, x[0]==a, x'[0]==b}, x[t], t]] then compare that with x[t]==Exp[m*t] to see what initial conditions can give you that. $\endgroup$ – Bill Oct 11 '18 at 18:55
  • $\begingroup$ thanks, but in my exercise i have to first of all find this parameter m and then solve differential equation, i though it may not like my teacher $\endgroup$ – Ben Oct 11 '18 at 19:01
  • $\begingroup$ Hummm.. That's a weird one. How about you assign f=Exp[m*t] without having assigned any prior value to m or t. Then you use the D function to find D[f,t] and D[f,{t,2}] and then you substitute all those into your (2 t+1) x''[t]+2(2 t-1)x'[t]-8 x[t]==0 Then could you think how you might use Solve on that? $\endgroup$ – Bill Oct 11 '18 at 20:21
  • $\begingroup$ I dont no how to use Solve, but I used help and thought something that.... fun[t_] := Exp[m t] , Dfun = D[fun[t], t],DDfun = D[Dfun, t],form = ForAll[{t}, (2 t + 1) DDfun + 2 (2 t - 1) Dfun - 8 fun[t] == 0],Resolve[form, Reals] thanks all of you $\endgroup$ – Ben Oct 11 '18 at 20:54
  • $\begingroup$ Interesting method you found. Solve[(2 t+1) DDfun+2 (2 t-1) Dfun-8 fun[t]==0,m] Then carefully check any result from MMA to make certain it is a real solution and not just a glitch. $\endgroup$ – Bill Oct 11 '18 at 21:56

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