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?

restart

with(PDEtools)

with(LinearAlgebra)

NULL

with(SolveTools)

undeclare(prime)

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

(1)

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

_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.

 

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)

n := 2

2

(3)

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)

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)

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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