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This question already has an answer here:

I want to solve the following differential equation with mathematica:

 α β * w''''''[ξ] + 
   ( 1 + α - p ) * w''''[ξ] + p/β * w''[ξ] = 0

The answer seems at first appearance really complicated.

w[ξ] -> 
 2 α β (-((
      E^((Sqrt[-((-1 + p - α + Sqrt[
          p^2 + 2 p (-1 + α) + (1 + α)^2])/(α \
β))] ξ)/Sqrt[2]) C[1])/(-1 + p - α + Sqrt[
       p^2 + 2 p (-1 + α) + (1 + α)^2])) - (
     E^(-((Sqrt[-((-1 + p - α + Sqrt[
          p^2 + 2 p (-1 + α) + (1 + α)^2])/(α \
β))] ξ)/Sqrt[2])) C[2])/(-1 + p - α + Sqrt[
      p^2 + 2 p (-1 + α) + (1 + α)^2]) + (
     E^((Sqrt[(
       1 - p + α + Sqrt[
        p^2 + 2 p (-1 + α) + (1 + α)^2])/(α \
β)] ξ)/Sqrt[2]) C[3])/(
     1 - p + α + Sqrt[
      p^2 + 2 p (-1 + α) + (1 + α)^2]) + (
     E^(-((Sqrt[(
        1 - p + α + Sqrt[
         p^2 + 2 p (-1 + α) + (1 + α)^2])/(α \
β)] ξ)/Sqrt[2])) C[4])/(
     1 - p + α + Sqrt[
      p^2 + 2 p (-1 + α) + (1 + α)^2])) + 
  C[5] + ξ C[6]

So now I want to simplify the result above. I am sure that the result can be simplified with some own defined expressions like:

a1 = Sqrt[p^2 + 2 p (-1 + α) + (1 + α)^2]

I have encountered similar differential equations with confusing solutions. I would really appreciate if someone could help me to solve this problem.

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marked as duplicate by Michael E2, bbgodfrey, Bob Hanlon, Jens, Mr.Wizard Feb 15 '15 at 22:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ Probable duplicate: (3822) (please see the many links in my answer there) $\endgroup$ – Mr.Wizard Feb 3 '15 at 10:58
  • $\begingroup$ Note that your original differential equation (which should use == rather than =) is actually a 4th order linear differential equation in v = w''. So you may want to try first simplifying the solution v[\[Xi]] of that 4th order equation before proceeding to integrate twice to come down to the original w. $\endgroup$ – murray Feb 3 '15 at 15:06
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This is the right-hand part of your solution (taken from your post above):

    expr = 2 α β (-((E^((Sqrt[-((-1 + p - α + 
                    Sqrt[p^2 + 
                    2 p (-1 + α) + (1 + \
α)^2])/(α β))] ξ)/Sqrt[2]) C[1])/(-1 + 
           p - α + 
           Sqrt[p^2 + 
             2 p (-1 + α) + (1 + α)^2])) - \
(E^(-((Sqrt[-((-1 + p - α + 
                    Sqrt[p^2 + 
                    2 p (-1 + α) + (1 + \
α)^2])/(α β))] ξ)/Sqrt[2])) C[2])/(-1 + 
         p - α + 
         Sqrt[p^2 + 
           2 p (-1 + α) + (1 + α)^2]) + (E^((Sqrt[(1 - 
                  p + α + 
                  Sqrt[p^2 + 
                    2 p (-1 + α) + (1 + \
α)^2])/(α β)] ξ)/Sqrt[2]) C[3])/(1 - 
         p + α + 
         Sqrt[p^2 + 
           2 p (-1 + α) + (1 + α)^2]) + (E^(-((Sqrt[(1 -
                     p + α + 
                    Sqrt[p^2 + 
                    2 p (-1 + α) + (1 + \
α)^2])/(α β)] ξ)/Sqrt[2])) C[4])/(1 - 
         p + α + 
         Sqrt[p^2 + 2 p (-1 + α) + (1 + α)^2])) + 
   C[5] + ξ C[6];

Try the following:

expr /. p^2 + 2 p (-1 + α) + (1 + α)^2 -> a1^2 /. 
 p -> 1 + Sqrt[a1^2] - b1 + α

where b1=1 + Sqrt[a1^2] - p + α. The result is somewhat more simple:

 2 α β (-((
     E^((Sqrt[-((2 Sqrt[a1^2] - b1)/(α β))] ξ)/Sqrt[
      2]) C[1])/(2 Sqrt[a1^2] - b1)) - (
    E^(-((Sqrt[-((2 Sqrt[a1^2] - b1)/(α β))] ξ)/Sqrt[
      2])) C[2])/(2 Sqrt[a1^2] - b1) + (
    E^((Sqrt[b1/(α β)] ξ)/Sqrt[2]) C[3])/b1 + (
    E^(-((Sqrt[b1/(α β)] ξ)/Sqrt[2])) C[4])/b1) + 
 C[5] + ξ C[6]

It will look much more readable, when obtained in Mma.

Mathematica graphics

Have fun!

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