Solving a Diff. Eq. Involving Primitive of sen x^3

[esmeco]

Homework Statement

I'm stuck trying to solve a differential equation at the point i need to calculate the primitive of sen x^3

Homework Equations

The Attempt at a Solution

I've thought on primitives by parts but I don't know how will I do it...

 

[Dick]

I don't think you have any hope of expressing that in terms of elementary functions.

 

[esmeco]

So you say it's not possible to solve that primitive??That is strange...maybe it was the books' author fault...

 

[Dick]

If by primitive you mean integral, yes. Maybe you could post your differential equation and your attempt to solve it. That step might not be necessary.

 

[esmeco]

The differential equation is: y' + (2/x)*y=cuberooty*sen x^3

 

[Dick]

What you mean on the right hand side is massively unclear. Is it:

y^{1/3} sin(x^3)

 

[esmeco]

well the y part you've wrote is ok but my doubt is in the x^3,because there aren't any parenthesis in my textbook so I don't know if they meant sin (x^3) or (sin x)^3!

 

[quantumdude]

Usually, \sin x^3=\sin(x^3), whereas \sin^3 x=sin^3(x). If your book doesn't use parentheses, try to figure out what is meant by context (as in, say, an example problem).

 

[esmeco]

It only says to solve the differential equation...And we have to use the Bernoulli formula to solve it...But if it's put like sin x^3 I don't know how to solve it...

 

[Dick]

You do want to use the Bernoulli form to start solving it. Do this change of variables first and show what you get. What you have to integrate in the end will not be just sin(x^3).

 

[esmeco]

What I have to integrate in the end is: (2/3)*sen x^3 *4x

 

[Dick]

Why do you keep saying "sen"? I'm not sure I believe that that is what you really have to integrate - but you're not showing any intermediate work, so I can't really comment. But even so, you don't have to integrate sin(x^3). It's already something else.

 

[esmeco]

I(x)= e^(primitive(4/3x))=e^4/3*ln(3x)=4x

Z=y^2/3
Z'=2/3*(y^-1/3)*y'


y'*(y^-1/3) + (2/x)*y^2/3=sin x^3 <=> 3/2*Z' + 2/x*Z=sin x^3 <=>

Z' + (4/3x)*Z=2/3*sin x^3

 

[Dick]

I'm not sure what rules you are using on I(x), but I don't think they are right. Try using a*ln(x)=ln(x^a).