You have a list of 828 sublists. Each sublist contains [0,0] plus two other points. You do a solve which requires four points. How are the four points obtained from the sublists? Which sublists did you "select by hand"?

Maple 17 is able to solve these with arbitrary u(t). It expresses the answers with integrals of u(t) times other functions.

Maple 17 is able to solve these with arbitrary u(t). It expresses the answers with integrals of u(t) times other functions.

The integrating factor that the Asker expects is what is called the "integrating factor" in a Calculus II course and also in the Wikipedia article "Integrating factor" and the MathWorld article "Integrating factor".

It seems that the result of DEtools:-intfactor depends on the form of the equation rather than on it's algebraic content. Using the same ode as the Asker:

ode1:= solve(ode, {diff(y(t),t)})[]:
intfactor(ode1);

which is the result that the Asker expected.

mutest(%, ode1);

                             0

Also note  what dsolve returns:

dsolve(ode);

Also compare

intfactor(f(t)*diff(y(t),t)+g(t)*y(t) = 0);

with

intfactor(diff(y(t),t)+g(t)/f(t)*y(t) = 0);

The integrating factor that the Asker expects is what is called the "integrating factor" in a Calculus II course and also in the Wikipedia article "Integrating factor" and the MathWorld article "Integrating factor".

It seems that the result of DEtools:-intfactor depends on the form of the equation rather than on it's algebraic content. Using the same ode as the Asker:

ode1:= solve(ode, {diff(y(t),t)})[]:
intfactor(ode1);

which is the result that the Asker expected.

mutest(%, ode1);

                             0

Also note  what dsolve returns:

dsolve(ode);

Also compare

intfactor(f(t)*diff(y(t),t)+g(t)*y(t) = 0);

with

intfactor(diff(y(t),t)+g(t)/f(t)*y(t) = 0);

It's not clear to me what you mean. Could you write some code, even though it may be wrong? Note that you can make substitutions for parameters inside procedures:

f:= proc(x) m*x end proc:
g:= subs(m=2, eval(f));

The problem on the left is a Ramanujan problem, and it is discussed in the Wikipedia article "Nested radicals." The convergence to 3 is very rapid. Fifty-one terms is enough to get 15 digits:

V:= 1.0:
for n from 52 by -1 to 2 do
     V:= sqrt(1.+n*V)
end do:
V;
                        3.00000000000000

@Heeka I don't know about it. But a Google search on "Homotopy perturbation method" (in quotes) turns up 407,000 hits. Note the correct spelling of "homotopy".

@Heeka I don't know about it. But a Google search on "Homotopy perturbation method" (in quotes) turns up 407,000 hits. Note the correct spelling of "homotopy".

When the Asker uploads the Mathematica Notebook, it should be in plain text form.

When the Asker uploads the Mathematica Notebook, it should be in plain text form.

@digerdiga Your code above works for me (if I put a semicolon after forget(x)). Try giving a restart command.

@digerdiga Your code above works for me (if I put a semicolon after forget(x)). Try giving a restart command.

@Adri van der Meer Not a bug. There is a subtle distinction between sqrt(x) and x^(1/2): The former checks whether x is a perfect square and the latter does not. Since 234256 is a perfect square, the computation with sqrt reduces to pure floating point.

@Adri van der Meer Not a bug. There is a subtle distinction between sqrt(x) and x^(1/2): The former checks whether x is a perfect square and the latter does not. Since 234256 is a perfect square, the computation with sqrt reduces to pure floating point.