Question: How do complicated substitution in ode?

i try to get same result by substituation but i don't know what is mistake after i take second derivative is wronge i don't know how get same result as in paper did can anyone help  to calculate this part is not hard but is complicated ,How calculated second derivative and put in our ode to get the parameters?

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)

declare(Omega(x, y, t)); declare(U(xi)); declare(u(x, y, t)); declare(Q(xi)); declare(V(xi)); declare(W(xi)); declare(f(xi))

Omega(x, y, t)*`will now be displayed as`*Omega

 

U(xi)*`will now be displayed as`*U

 

u(x, y, t)*`will now be displayed as`*u

 

Q(xi)*`will now be displayed as`*Q

 

V(xi)*`will now be displayed as`*V

 

W(xi)*`will now be displayed as`*W

 

f(xi)*`will now be displayed as`*f

(2)

NULL

ode := -delta*(diff(diff(U(xi), xi), xi))+U(xi)*(w^2-gamma*U(xi)-beta-alpha) = 0

-delta*(diff(diff(U(xi), xi), xi))+U(xi)*(w^2-gamma*U(xi)-beta-alpha) = 0

(3)

ode1 := -delta*(diff(diff(f(xi), xi), xi))+f(xi)*(w^2-gamma*f(xi)-beta-alpha) = 0

-delta*(diff(diff(f(xi), xi), xi))+f(xi)*(w^2-gamma*f(xi)-beta-alpha) = 0

(4)

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

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

(5)

S := U(f(xi)) = sum(cos(f(xi))^(i-1)*(B[i]*sin(f(xi))+A[i]*cos(f(xi))), i = 1 .. n)+A[0]

U(f(xi)) = sum(cos(f(xi))^(i-1)*(B[i]*sin(f(xi))+A[i]*cos(f(xi))), i = 1 .. n)+A[0]

(6)

``

n := 2

2

(7)

eval(ode1, S)

-delta*(diff(diff(f(xi), xi), xi))+f(xi)*(w^2-gamma*f(xi)-beta-alpha) = 0

(8)

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