For this integro-differential equation,

Equation:= int[y'(x)* (x^2)/[(x^2)-1],x)  =  (int[sqrt(y(x)])^(-2/3)

Maple is able to obtain an exact intrinsic solution

from which an exact solution can be extracted, namely,

ExtrinsicSolution:= y(x) = sqrt(3)*(-8*_C1*x^(8/3) + 12*x^2 - 3)^(3/4)

My question concerns how was this solution obtained.

Even more, specifically, 'odeadvisor' suggests converting the

equation in question to the form

ode:= y = G(x,diff(y(x),x));

However, I cannot reconcile how this can be applied to an equation which

contains two integrals. (Regretably, I am not able to directly, attach my

Maple worksheet directly on to this sheet). The situation is that after

applying 'dsolve' to the above 'Equation', Maple comes back with an

intrinsic solution which can was used to obtain the 'ExtrinsicSolution' in 

the above.  So it is the missing steps between applyingthe dsolve command

to Equation and the intrinsic solution which MS provides which, in turn, leads

to the 'Extrinsic Solution' above. I would greatly, appreciate if anyone can 

fill in the missing steps.

For the following Equation:

Equation := int(diff(u(x), x)*v(x), x) = int(u(x)^(1/2)*v(x), x)^(-2/3);
Maplesoft finds the following solution:

Solution1:=3/4*u(x)^(4/3) + 2/3*u(x)^(5/6)*Intat(1/(sqrt(u(x))*Int(v(_b), _b))^(5/3), _b = x) + _C1 = 0

or , which I believe as an alternative, can be written as

Solution2:=3/4*u(x)^(4/3) + 2/3*u(x)^(5/6)*Int(1/(sqrt(u(x))*Int(v(x),x))^(5/3) +_C1=0

My question is how did Maple arrive at 'Solution1' from 'Equation'? In other words, can someone fill

in the steps between 'Equation'  and 'Solution1'? Or even, prove that Solution 1 is a valid solution to Equation.

Plugging the Solution1 into Equation, did not clearly demonstate the validity of the solution (to me at least)

Unfortunately, I am still unable to post the corresponding Maplesoft worksheet onto this post.

Equation:=int(y'(x)*w(x)) = [int([y(x)^(1/2)]*w(x))]^(-2/3) When plugging this Equation into dsolve, Maple provides the following implicit solution: Solution:= [3*y(x)^(4/3)]/4 + int[(2*y(x)^(5/6)/[3*int([(y(x)^(1/2))*w(x)]/[3*int[(y(x)^(1/2))*w(x)]^(5/3) When going to odeadvisor, the suggestion was to first convert to the form y = G(x,y'(x) and then utilize the method of 'patterns', which I could not apply to this equation. If anyone can fill in the steps between 'Equation' and 'Solution', it would be greatly, appreciated. P.S.Sadly, I am unable to attach the actual Maplesoft worksheet.

How could one reduce this equation int(x^2*diff(y(x), x)/(x^2 - 1), x) = int(y(x)^(1/2), x)^(-2/3)down to a first order ordinary differential equation? Maple can solve this equation, namely, 9*(1/x^(5/3) - 1/x^(11/3))*x*(sqrt(y(x))*x)^(8/3)/(8*(x^2 - 1)) - 3*x*(4*x^2 - 1)*(1/x^(5/3) - 1/x^(11/3))/(8*(x - 1)*(x + 1)) + _C1 = 0 , however I could not understand how this equation was arrived at, leading me to go to 'odeadvisor', which responded with y = G(x,y'(x)) labelled as the 'patterns' method, which appears to require a first order ode - that is why I raised the reduction question.

How did Maple arrive at the following, implicit solution,

MSimplicitesolution := (-9*y(x)*(-y(x)*x^2)^(1/3)*6^(2/3)*x^2 + 16*_C1*x^(8/3) - 24*x^2 + 6)/(16*x^(8/3)) = 0, 

to the following 'ode',

ODE:= (2/3)*(int[(y'(x)*(x^2)/((x^2) -1)]) =int(-sqrt [2*y(x)])?

'odeadvisor' suggested y=G(x,y'(x), but I could not see how this could be implimented with this equation.

(Unfortunately, I am unable to download the Maple worksheet onto this sheet at this time.  Any assitance

would be appreciated.