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Mathematica solves this but I want the solution just as Ln[..]?!!

DSolve[y'[t] == 3/2 a*y[t] ((y[t] + b)^2 - f), y[t], t]

thanks

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1 Answer 1

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This is a separable ODE of the first order. Let us make one step by hand and obtain

t == Integrate[1/(3/2 a*y[t] ((y[t] + b)^2 - f)) /. y[t] -> z, z] /. z -> y[t]

t+ C[1]==(2 (-((2 b ArcTanh[(b+y[t])/Sqrt[f]])/Sqrt[f])-2 Log[y[t]]+Log[-f+(b+y[t])^2]))/(3 a (-2 b^2+2 f))

instead of

InverseFunction[(-((2 b ArcTanh[(b + #1)/Sqrt[f]])/Sqrt[f]) - 2 Log[#1] + Log[b^2 - f + 2 b #1 + #1^2])/(-2 b^2 + 2 f) &][( 3 a t)/2 + C[1]]

produced by

DSolve[y'[t] == 3/2 a*y[t] ((y[t] + b)^2 - f), y[t], t]

Hope you feel the difference.

PS.

t + C[1] == (2 (-((2 b ArcTanh[(b + y[t])/Sqrt[f]])/Sqrt[f]) - 
   2 Log[y[t]] + Log[-f + (b + y[t])^2]))/(3 a (-2 b^2 + 
   2 f)) // TrigToExp

t + C[1] == (1/(3 a (-2 b^2 + 2 f))) 2 (-2 Log[y[t]] - ( b (-Log[1 - (b + y[t])/Sqrt[f]] + Log[1 + (b + y[t])/Sqrt[f]]))/ Sqrt[f] + Log[-f + (b + y[t])^2])

gets rid of ArcTanh.

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  • $\begingroup$ I would like to have the answer all on based Log[] but I have this 2 b ArcTanh[(b + y[t])/Sqrt[f]])/Sqrt[f] $\endgroup$ May 4 at 15:25
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    $\begingroup$ @ArminSharafi Try applying the function TrigToExp to get rid of ArcTanh. $\endgroup$
    – march
    May 4 at 16:41
  • $\begingroup$ I saw your comment thank you very much,. $\endgroup$ May 4 at 18:25

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