I am trying to solve

-D[((α + γ/2)*a + β/(2*a) - R*a^3)*p[a], a] + 0.5*D[(γ*a^2 + β)*p[a], {a, 2}] == 0

with DSolve, but it does not work. If I let R = 0, then I am able to obtain an explicit solution for p[a].

    -(-3 a^2 R + α - β/(2 a^2) + γ/2) p[a] - 
    (-a^3 R + β/(2 a) + a (α + γ/2)) Derivative[1][p][a] + 
    0.5 (2 γ p[a] + 4 a γ Derivative[1][p][a] + (β + a^2 γ) (p^′′)[a]) == 0, 
  p[a], a]

Can someone kindly help me on this.


1 Answer 1


I evaluated

DSolve[-2 D[((α + γ/2) a + β/(2 a) - R a^3)*p[a], a] + D[(γ a^2 + β) p[a], {a, 2}] == 0, 
  p[a], a]

which is equivalent to the first formulation of your problem, but which eliminates the approximate coefficient (i.e., 0.5). Mathematica V9.0.1 returned

{{p[a] -> 
  a*E^(-((a^2*R)/γ) + ((R*β + α*γ - γ^2)*Log[β + a^2*γ])/γ^2)*C[1] + 
  a*E^(-((a^2*R)/γ) + ((R*β + α*γ - γ^2)*Log[β + a^2*γ])/γ^2)*C[2]*
    Integrate[E^((R*K[1]^2)/γ - ((R*β + α*γ)*Log[β + γ*K[1]^2])/γ^2)/K[1], 
      {K[1], 1, a}]}}

Does this result work for you, or do you find it unsatisfactory? If unsatisfactory, please explain why.

  • $\begingroup$ thanks a lot for this....I did not know that I needed to get rid of the 0.5. Regards! $\endgroup$
    – Stoc
    Jun 12, 2014 at 15:00
  • $\begingroup$ although it should be possible to solve this ODE by hand...is it possible that Mathematica can show me the steps involved? As wolframalpha does? $\endgroup$
    – Stoc
    Jun 12, 2014 at 15:16
  • $\begingroup$ @Stoc. I don't know a way to instruct Mathematica to show how it gets its results in a step-by-step way. My understanding is that its methods are not much like the ones humans use and would be unreadable to us humans. I have heard that Wolfram|Alpha can produce step-by-step results, but have no personal experience with that. $\endgroup$
    – m_goldberg
    Jun 12, 2014 at 20:16
  • 1
    $\begingroup$ @Stoc. I think I should mention that it's not so much a matter of "need[ing] to get rid of the 0.5", but of replacing any approximate numbers with integers or rationals. Changing 0.5 to 1/2 would have worked just as well as multiplying by integer 2. $\endgroup$
    – m_goldberg
    Jun 12, 2014 at 20:22

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