Give more details maybe someone can answer this?

Root two is an irrational number and you take this value for a
If you then assume that x is a real number. 
Then what will maple calculate for the function values ?

@scallopedpancake 
Setup screen ? in Physics package 

As a system of odes via  ODEsteps : lambda and mu  , for symbolic solving  there are no solutionsteps , but for lambda and mu = 1 there is a solution
Note : output odesteps not correct ?


 

Download Find_ODE_solution_with_condition_parameter_als_syteem_via_odesteps_26-11-2024.mw

@dharr  "avoids square roots of negative numbers in practice" . This is calculated in the complex number system.
 

restart;

# Define the coupled differential equations
de1 := diff(F(eta), eta) = lambda * G(eta):
de2 := diff(G(eta), eta) = mu * F(eta):

# Solve the differential equations (general solution)
sol := dsolve({de1, de2}, {F(eta), G(eta)}):
sol_hyperbolic := convert(sol, trigh):  # Hyperbolic form

# General solution (hyperbolic, valid for all lambda and mu)
print("=== General Solution (valid for all lambda and mu) ==="):
print(sol_hyperbolic);

# Alternative solution for lambda * mu < 0 (explicit complex root)
assume(lambda < 0, mu > 0):  # Assume lambda * mu < 0
complexRoot := simplify(sqrt(lambda * mu)):  # Explicitly compute complex root
complexRoot_imag := I * sqrt(-lambda * mu):  # Write root in imaginary form

# Replace hyperbolic functions with trigonometric functions
sol_trig := subs(sqrt(lambda * mu) = complexRoot_imag, sol_hyperbolic):
sol_trig_converted := simplify(sol_trig):
print("=== Alternative Solution (Trigonometric Form for lambda * mu < 0) ==="):
print(sol_trig_converted);

# Add information from the image: interpretation and cases
print("=== Interpretation of Cases ==="):
print("Case (I): Covers all combinations of lambda and mu signs using hyperbolic functions.");
print("Case (II): If lambda * mu < 0, the solution can be rewritten in trigonometric form.");
print("Note: Hyperbolic functions naturally handle both positive and negative values of lambda * mu.");

# Validate both solutions with odetest
validation_hyperbolic := odetest(sol_hyperbolic, [de1, de2]):
validation_trig := odetest(sol_trig_converted, [de1, de2]):

print("=== Validation of Solutions ==="):
print("Validation of the hyperbolic solution:"):
print(validation_hyperbolic):  # Result must be 0
print("Validation of the trigonometric solution (lambda * mu < 0):"):
print(validation_trig);  # Result must be 0

No complex numbers anymore or do you mean something else ?

@Carl Love 

Yes, some explanation of the code can never hurt in my opinion.

But to really understand maple code better, the AI is also a good helper, but paid via ChatGPTplus
Unfortunately the Maple coding expert does not seem to be around anymore, so it will be Wolfram or yet other chats .

@Aung 

Then you should rewrite a general expression of the current integral expression into a general form of an integral by parts ?
 =   ( with partiele integration )

Both expressions i let check this  with  parameters and they seem to be numeric equal 


 

@Aung 

This is another methode from acer of solving your expression than dharr did for your expression.

acer methode :
"Do you need a fully programmatic way to transform the integral of a sum into the sum of integrals? For example, programmatically extracting and re-using not just the integrand but also the ranges of integration and summation".

Suppose you take the first term of the sum and solve this integral by parts : what to do with this expression?

@dharr 

I'm not a fan of your code, which only shows maple commands.
Keep in mind , that someone is asking for your help and needs some more explanation. 

Make "free" a variable 

x := 5;  # Assigning a value to x
x := 'x';# Releasing the variable by resetting it to a symbol


                             x := 5

                             x := x

@salim-barzani 

??

paramter_alphao_etclniet_lin_ode_9-11-2024-deel_2.mw

@salim-barzani 
?
paramater_bepaling_niet_lin_ode_9-11-2024.mw

@janhardo 

Download polynoom_adomian_uitvoerprocedure_2_variabelen_in_tabel_7-11-2024.mw

seems that output:
[u[0]*ux[0]*uxx[0], u[0]*ux[0]*uxx[1] + u[0]*ux[1]*uxx[0] + u[1]*ux[0]*uxx[0], u[0]*ux[0]*uxx[2] + u[0]*ux[1]*uxx[1] + u[0]*ux[2]*uxx[0] + u[1]*ux[0]*uxx[1] + u[1]*ux[1]*uxx[0] + u[2]*ux[0]*uxx[0]]
could be correct in a different notation only 
to check...

L := [u[0]*ux[0]*uxx[0], 
      u[0]*ux[0]*uxx[1] + u[0]*ux[1]*uxx[0] + u[1]*ux[0]*uxx[0], 
      u[0]*ux[0]*uxx[2] + u[0]*ux[1]*uxx[1] + u[0]*ux[2]*uxx[0] + 
      u[1]*ux[0]*uxx[1] + u[1]*ux[1]*uxx[0] + u[2]*ux[0]*uxx[0]]:

# Toewijzen van elk element aan een dynamische naam L1, L2, ...
for i from 1 to nops(L) do
    assign(cat(L, i), L[i]);
end do:

# Weergeven van de variabelen en hun waarden
for i from 1 to nops(L) do
    print( polynome(i-1)= eval(cat(L, i)));
end do:

                polynome(0) = u[0] ux[0] uxx[0]

      polynome(1) = u[0] ux[0] uxx[1] + u[0] ux[1] uxx[0]

         + u[1] ux[0] uxx[0]

 polynome(2) = u[0] ux[0] uxx[2] + u[0] ux[1] uxx[1]

    + u[0] ux[2] uxx[0] + u[1] ux[0] uxx[1] + u[1] ux[1] uxx[0]

    + u[2] ux[0] uxx[0]