# ExpToTrig transforms solution to 4th order ODE into unwanted form

Mathematica gives the solution of the second order differential equation

 DSolve[a y''[x] + b*y[x] == 0, y[x], x]


in trigonometric form

{{y[x] -> C[1] Cos[(Sqrt[b] x)/Sqrt[a]] + C[2] Sin[(Sqrt[b] x)/Sqrt[a]]}}


In the fourth order case, Mathematica gives the solution in exponential form, which I tried to convert into trigonometric form

  ExpToTrig[DSolve[a y''''[x] + b*y[x] == 0, y[x], x]]

 {{y[x] ->
C[1] Cos[((1 + I) b^(1/4) x)/(Sqrt[2] a^(1/4))] +
C[3] Cos[((1 + I) b^(1/4) x)/(Sqrt[2] a^(1/4))] +
C[2] Cosh[((1 + I) b^(1/4) x)/(Sqrt[2] a^(1/4))] +
C[4] Cosh[((1 + I) b^(1/4) x)/(Sqrt[2] a^(1/4))] +
I C[1] Sin[((1 + I) b^(1/4) x)/(Sqrt[2] a^(1/4))] -
I C[3] Sin[((1 + I) b^(1/4) x)/(Sqrt[2] a^(1/4))] -
C[2] Sinh[((1 + I) b^(1/4) x)/(Sqrt[2] a^(1/4))] +
C[4] Sinh[((1 + I) b^(1/4) x)/(Sqrt[2] a^(1/4))]}}


The problem is that I get a solution with hyperbolic functions, but I know that solution can be represented in purely trigonometric form. Another problem is that I get an imaginary part from multiplication of the constants I C[1] and I C[3], which I wouldn't have a with trigonometric function solution. How can I transform the solution to a form having only trigonometric functions?

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How about HoldForm@Evaluate@DSolve[a y''''[x] + b*y[x] == 0, y[x], x] /. E^x_ :> (Cos[-I x] + I Sin[-I x]) ? – xzczd Mar 26 '14 at 2:42