Simplify this equation in terms of function and coefficients?

Posted 1 month ago
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 I have two fourth order differential equations and I have solved both equations on Dsolve to get the final solution but it gives me a very large output. I need the result in the form of function (Y) and constant coefficient (let say p1,p2,p3 and so on). So how can I transform it into a simplified form? Here are the equations. X''''[Y] - A6*X''[Y] + (B1*Bi)/(k*Um)*(Upm) - B1/(k*Um)*(D[C1*Sinh[A0*Y] + C2*Cosh[A0*Y] + Da, {Y, 2}]) - ( Bi*Br*Da)/( k*Um^2)*(1/Da*(Upm)^2 + 1/\[Epsilon]*(D[ C1*Sinh[A0*Y] + C2*Cosh[A0*Y] + Da, {Y, 1}])^2) + (Br*Da)/( Um^2*k)*(1/ Da*(D[(C1*Sinh[A0*Y] + C2*Cosh[A0*Y] + Da)^2, {Y, 2}]) + 1/\[Epsilon]*(D[(A0 (C1 Cosh[A0 Y] + C2 Sinh[A0 Y]))^2, {Y, 2}])) == 0 Z''''[Y] - A6*Z''[Y] + (B1*Bi)/(k*Um)*(Upm) - (Bi*Br*Da)/( k*Um^2)*(1/Da*(Upm)^2 + 1/\[Epsilon]*(D[ C1*Sinh[A0*Y] + C2*Cosh[A0*Y] + Da, {Y, 1}])^2) == 0 Boundary Conditions X''[0] == B4, Z''[0] == 0, X[0] == 0, Z[0] == 0, X''[t] == B3, Z''[t] == 0, X[t] == Z[t], B2 == -k*X'[t] - Z'[t] 
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Posted 1 month ago
 If sol is the formula for X[Y], for example, I would try this as a start: Collect[TrigToExp@Numerator[Together[sol]] /. E^u_ :> E^Collect[u, Y], Y^_.*E^(_. + _*Y), Style[Simplify[#], RandomColor[]] &] It needs more work, but I have to go now.
 I tried to simplify the equation further but still don't get the simplified result. Can you help me further Please? Here is my code eqn1 = X''''[Y] - A6*X''[Y] - ( E^(-2 A0 Y) (A0^2 (E^(A0 Y) G1 + E^(3 A0 Y) G2 + 4 G3 + 4 E^(4 A0 Y) G4) - Bi (E^(A0 Y) G1 + E^(3 A0 Y) G2 + G3 + E^(4 A0 Y) G4 + E^(2 A0 Y) G5)))/k == 0; eqn2 = Z''''[Y] - A6*Z''[Y] + Bi/k*(G1 E^(-A0 Y) + G2 E^(A0 Y) + G3 E^(-2 A0 Y) + G4 E^(2 A0 Y) + G5) == 0; system = {eqn1, eqn2, X''[0] == B4, Z''[0] == 0, X[0] == 0, Z[0] == 0, X''[t] == B3, Z''[t] == 0, X[t] == Z[t], B2 == -k*X'[t] - Z'[t]}; DSolveValue[system, {X[Y], Z[Y]}, Y]