Question: How find parameters in in equation involving hyperbolic function?

the most paper use another function to get the result and then do substitute i try to get by the way of them but i fail so how about if we can get the results in direct function there is any way for finding thus parameter after substitution of our function in ode?

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restart

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with(PDEtools)

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with(LinearAlgebra)

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NULL

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with(SolveTools)

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undeclare(prime)

`There is no more prime differentiation variable; all derivatives will be displayed as indexed functions`

(1)
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"          with(Student[ODEs][Solve]):"

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_local(gamma)

Warning, A new binding for the name `gamma` has been created. The global instance of this name is still accessible using the :- prefix, :-`gamma`.  See ?protect for details.

 
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ode := beta*U(xi)^2*c^2+(-alpha*c^2+1)*U(xi)+mu^2*c^2*(diff(diff(U(xi), xi), xi)) = 0

beta*U(xi)^2*c^2+(-alpha*c^2+1)*U(xi)+mu^2*c^2*(diff(diff(U(xi), xi), xi)) = 0

(2)
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n := 2

2

(3)
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F := U(xi) = sum(tanh(xi)^(i-1)*(B[i]*sech(xi)+A[i]*tanh(xi)), i = 1 .. n)+A[0]

U(xi) = B[1]*sech(xi)+A[1]*tanh(xi)+tanh(xi)*(B[2]*sech(xi)+A[2]*tanh(xi))+A[0]

(4)
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K1 := eval(ode, F)

beta*(B[1]*sech(xi)+A[1]*tanh(xi)+tanh(xi)*(B[2]*sech(xi)+A[2]*tanh(xi))+A[0])^2*c^2+(-alpha*c^2+1)*(B[1]*sech(xi)+A[1]*tanh(xi)+tanh(xi)*(B[2]*sech(xi)+A[2]*tanh(xi))+A[0])+mu^2*c^2*(B[1]*sech(xi)*tanh(xi)^2-B[1]*sech(xi)*(1-tanh(xi)^2)-2*A[1]*tanh(xi)*(1-tanh(xi)^2)-2*tanh(xi)*(1-tanh(xi)^2)*(B[2]*sech(xi)+A[2]*tanh(xi))+2*(1-tanh(xi)^2)*(-sech(xi)*tanh(xi)*B[2]+A[2]*(1-tanh(xi)^2))+tanh(xi)*(sech(xi)*tanh(xi)^2*B[2]-sech(xi)*(1-tanh(xi)^2)*B[2]-2*A[2]*tanh(xi)*(1-tanh(xi)^2))) = 0

(5)
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solve(identity(K1, {xi}), {A[0], A[1], A[2], B[1], B[2]})

Error, (in unknown) incorrect use of identity(<expr>,<name>)

 

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