# Solve an equation for different values

I was trying to solve this equation for different value of a constant (a) but the solution is not different for those values?!! b and m are positive values.

Clear["Global*"]
eqns = {(y'[t]/ y[t])^2 == b / y[t]^(3m) + a/3};
Simplify[DSolve[eqns, y[t], t], Assumptions -> {a > 0}]
Simplify[DSolve[eqns, y[t], t], Assumptions -> {a < 0}]

• Why do you think it is not working? You clearly do get a solution, but Mathematica just cannot express it explicitly so it is written as a InverseFunction. If there exists an explicit solution, you will probably need to do some manual transformations – see for example DSolve returning InverseFunction while I believe the answer isn't so complicated. Jan 10, 2023 at 14:38
• I'm sorry, you'r right but I want to know the manual transformations Jan 10, 2023 at 14:49
• Then please change the title and the content of your question accordingly :) Jan 10, 2023 at 15:31

You could try to solve the inverse problem

eqns = {(y'[t]/y[t])^2 == b/y[t]^(3 m) + a/3}
solys = Solve[eqns, y'[t]]

odes = Map[t'[y] == 1/y'[t] /. # /. y[t] -> y &, solys](* ode in t[y]:
{Derivative[1][t][y] == -(Sqrt[3]/Sqrt[a y^2 + 3 b y^(2 - 3 m)   ]),
Derivative[1][t][y] == Sqrt[3]/Sqrt[a y^2 + 3 b y^(2 - 3 m)]} *)

DSolve[odes[[1]], t, y]
(* {{t -> Function[{y}, -((2 y Sqrt[a + 3 b y^(-3 m)]ArcTanh[Sqrt[a + 3 b y^(-3 m)]/Sqrt[a]])/(Sqrt[3] Sqrt[a] m Sqrt[y^2 (a + 3 b y^(-3 m))])) + C[1]]}}*)

DSolve[odes[[2]], t, y]
(*{{t -> Function[{y}, (2 y Sqrt[a + 3 b y^(-3 m)]ArcTanh[Sqrt[a + 3 b y^(-3 m)]/Sqrt[a]])/(Sqrt[3] Sqrt[a] m Sqrt[y^2 (a + 3 b y^(-3 m))]) + C[1]]}}*)


Hope it helps!

• This is a good idea but I'm working on some condition that my solution will be Sin [] and Sinh[] ! Jan 10, 2023 at 17:06
• Like this Clear["Global*"] eqns = {(y'[t]/ y[t]) == Sqrt[b/ y[t]^(3m) + a/3]}; Simplify[DSolve[eqns, y[t], t], Assumptions -> {a > 0, b > 0, m > 1}] Jan 10, 2023 at 17:08