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I'm new to Mathematica and I don't understand why in the solution of the following ODE, the #1 in the pure function is not immediately replaced by the corresponding [t/4 + C[1]]. It seems as if the program wanted to compute the "InverseFunction" operator before inserting the bracket into the solution. I don't see why it should matter mathematically to replace the #1 by the bracket only after computing the inverse function.

Further, why is the InverseFunction not computed? I was expecting to get one branch together with the message that there might be other branches.

sol = DSolve[{a'[t] == 
1/2 (-(3/2) + Sqrt[(-(3/2) - a[t])^2 - 2 a[t]] - a[t])}, a[t], t]

{{a[t] -> InverseFunction[
 1/8 (-ArcSinh[(1 + 2 #1)/(2 Sqrt[2])] - 6 Log[#1] + 
     3 Log[9 + 2 #1 + 3 Sqrt[9 + 4 #1 + 4 #1^2]] - 2 #1 - Sqrt[
     9 + 4 #1 + 4 #1^2]) &][t/4 + C[1]]}}
share|improve this question
You will get a specific value once you specify the parameters, like in sol[[1, 1, 2]] /. {t -> 1., C[1] -> 2.}. – b.gatessucks Jan 9 '13 at 16:33
Thanks that works. I think the problem is that there simply is no closed form solution, which I would have liked a lot. And then of course it matters a lot whether you plug in f(t) into #1 before or after the inversion. My bad. – malumno Jan 9 '13 at 16:45
Good question, though--have an upvote. And welcome to the site! It's always nice when a self-described newbie takes the time to ask a coherent and properly formatted question. :) – Oleksandr R. Jan 9 '13 at 16:59

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