Joseph Poveromo

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17 years, 182 days

MaplePrimes Activity


These are replies submitted by Joseph Poveromo

@C_R 

Thank you, for addressing my equation, namely,  int(f(x)*y'(x),x) = [int(f(x)*sqrt(y(x)),x]^(-2/3).  Next, if there is a direct email where I can send the actual worksheet, as I was unable to paste the actual Maplesoft work sheet onto this reply.  Such an email would clarify what follows a great deal.  Nevertheless, I will do my best.

The burning question is how did Maplesoft solve it.  For this, I first went to 'with(DEtools)' on Maplesoft, which responded firstly, by saying that the equation had to put into the form y= G(x,y'(x)) and then, utilize the method of 'patterns'. When applying 'dsolve' to the equation the following implicit solution results,

                          (3/4)*y(x)^(4/3) + int{(2*y(x)^(5/6))/(3*([int(f(b)*sqrt(y(x)),b->x)]^(5/3))) +_C1=0

My interpretation of this equation is the following,

                                        y(x)=2*(sqrt(6)/9)*{2*[int(f(x),x))]^(-5/3),x]^(-5/3) -3*C1}^(3/4)

Again, if the steps between, the equation and the implicit solution can be filled in, I would be most grateful.

The Best,

Joe Poveromo

 

@Mariusz Iwaniuk  You are correct.  Thanks for pointing that out. The equation should read,

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

Equation := int(diff(y(x), x)*w(x), x) = 1/int(sqrt(y(x))*w(x), x)^(2/3);
Solution := 3/4*y(x)^(4/3) + int(2/3*y(x)^(5/6)/int(sqrt(y(x))*w(x), x)^(5/3), x) + _C1 = 0;
Again, the advice provided by 'odeadvisor', was to formulate the Equation to the form
y(x) = G(x,y'(x)), then utilize the 'patterns' method which I could not apply, therefore it is the missing steps between 'Equation' and 'Solution'

Sorry,

JJP

Your advice is well taken - Thanks. Incorporating it, I will give it a second shot - I hope you will too. Equation := int(diff(y(x), x)*w(x), x) = 1/int(sqrt(y(x))*w(x), x)^(2/3); Solution := 3/4*y(x)^(4/3) + int(2/3*y(x)^(5/6)/int(sqrt(y(x))*w(x), x)^(5/3), x) + _C1 = 0; Again, the advice provided by 'odeadvisor', was to formulate the Equation to the form y(x) = G(x,y'(x)), then utilize the 'patterns' method which I could not apply, therefore it is the missing steps between 'Equation' and 'Solution' Thanks Again, JJP

Corection: Sorry all, I left out an exponential factor on the left side of the equation. 

The Equation should read,

 (2/3)*[int(diff(y(x),x)*x^2/(x^2 -1),x)]^(-3/2) =int(-sqrt(2*y(x)),x). It is that exponential

factor of (-3/2) on the left side of the equation which I left out...JJP

@Preben Alsholm 

Thanks for taking the time. It is who made the error. The Equation should have been (2/3)*[int(diff(y(x),x)*x^2/(x^2 -1),x)]^(-3/2) =int(-sqrt(2*y(x)),x). It is that exponential factor of (-3/2) on the left side of the equation which I left out. Sorry

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