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Question: How this ODE will be zero-the result in not acceptable why?

Question:How this ODE will be zero-the result in not acceptable why?

salim-barzani 185 Maple 2024
ode homework differential-equations odetest
December 11 2024
0

this equation will be solve by changing variable but when  i found the function and substitute the ODE is not zero where  is mistake?

> 

restart
> 

with(PDEtools); _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.

  
> 

undeclare(prime)

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

 (1)
> 

declare(phi(x, t)); declare(psi(x, t)); declare(U(xi))

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

  

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

  

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

 (2)
> 

with(PDEtools)
> 

with(LinearAlgebra)
> 

NULL
> 

with(SolveTools)
> 

undeclare(prime)

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

 (3)
> 

ode := (diff(diff(U(xi), xi), xi))*lambda^2+(diff(diff(diff(diff(U(xi), xi), xi), xi), xi))*lambda*k^3-6*(diff(diff(U(xi), xi), xi))*k^2*(diff(U(xi), xi))*lambda = 0

(diff(diff(U(xi), xi), xi))*lambda^2+(diff(diff(diff(diff(U(xi), xi), xi), xi), xi))*lambda*k^3-6*(diff(diff(U(xi), xi), xi))*k^2*(diff(U(xi), xi))*lambda = 0

 (4)
> 

W := diff(U(xi), xi) = T(xi)

diff(U(xi), xi) = T(xi)

 (5)
> 

ode1 := lambda^2*T(xi)+lambda*k^3*(diff(diff(T(xi), xi), xi))-3*k^2*lambda*T(xi)^2 = 0

lambda^2*T(xi)+lambda*k^3*(diff(diff(T(xi), xi), xi))-3*k^2*lambda*T(xi)^2 = 0

 (6)
> 

K := T(xi) = A[0]+A[1]*(exp(2*xi)-1)/(exp(2*xi)+1)+A[2]*(exp(2*xi)-1)^2/(exp(2*xi)+1)^2+B[1]*(exp(2*xi)+1)/(exp(2*xi)-1)+B[2]*(exp(2*xi)+1)/(exp(2*xi)-1)

T(xi) = A[0]+A[1]*(exp(2*xi)-1)/(exp(2*xi)+1)+A[2]*(exp(2*xi)-1)^2/(exp(2*xi)+1)^2+B[1]*(exp(2*xi)+1)/(exp(2*xi)-1)+B[2]*(exp(2*xi)+1)/(exp(2*xi)-1)

 (7)
> 

case1 := [k = (1/2)*A[2], lambda = -(1/2)*A[2]^3, A[0] = -A[2], A[1] = 0, A[2] = A[2], B[1] = -B[2], B[2] = B[2]]

[k = (1/2)*A[2], lambda = -(1/2)*A[2]^3, A[0] = -A[2], A[1] = 0, A[2] = A[2], B[1] = -B[2], B[2] = B[2]]

 (8)
> 

F1 := subs(case1, K)

T(xi) = -A[2]+A[2]*(exp(2*xi)-1)^2/(exp(2*xi)+1)^2

 (9)
> 

F2 := subs(case1, ode1)

(1/4)*A[2]^6*T(xi)-(1/16)*A[2]^6*(diff(diff(T(xi), xi), xi))+(3/8)*A[2]^5*T(xi)^2 = 0

 (10)
> 

odetest(F1, F2)

0

 (11)
> 

subs(F1, W)

diff(U(xi), xi) = -A[2]+A[2]*(exp(2*xi)-1)^2/(exp(2*xi)+1)^2

 (12)
> 

E := map(int, diff(U(xi), xi) = -A[2]+A[2]*(exp(2*xi)-1)^2/(exp(2*xi)+1)^2, xi)

U(xi) = A[2]*((1/2)*ln(exp(2*xi))+2/(exp(2*xi)+1))-A[2]*xi

 (13)
> 

odetest(E, ode)

32*A[2]*exp(8*xi)*lambda*k^3/(exp(2*xi)+1)^5-352*A[2]*exp(6*xi)*lambda*k^3/(exp(2*xi)+1)^5+192*A[2]^2*exp(6*xi)*lambda*k^2/(exp(2*xi)+1)^5+8*A[2]*exp(8*xi)*lambda^2/(exp(2*xi)+1)^5+352*A[2]*exp(4*xi)*lambda*k^3/(exp(2*xi)+1)^5-192*A[2]^2*exp(4*xi)*lambda*k^2/(exp(2*xi)+1)^5+8*A[2]*exp(6*xi)*lambda^2/(exp(2*xi)+1)^5-32*A[2]*exp(2*xi)*lambda*k^3/(exp(2*xi)+1)^5-8*A[2]*exp(4*xi)*lambda^2/(exp(2*xi)+1)^5-8*A[2]*exp(2*xi)*lambda^2/(exp(2*xi)+1)^5

 (14)
 

NULL

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