Solving Bernoulli's DE: xy'+y+x^4y^4e^x=0

[fluidistic]

Homework Statement

I must solve xy'+y+x^4y^4e^x=0.

Homework Equations

Bernoulli's.

The Attempt at a Solution

I divided the original DE by x to get y'+y \left ( \frac{1}{x} \right )=-x^3e^xy^4.
Now let z=y^{-3} \Rightarrow z'=-3y^{-3}y'.
I then multiplied the DE by -3y^{-4} to reduces the DE to z'-\frac{3}{x}z=3x^3e^x. This is a first order linear DE so I should be able to solve it via the integrating factor method, however this doesn't work out for me.
The integrating factor is e^{\int -3 /x dx}=x^{-3}. So that the general solution of this DE (the z's one) should be z=-e^xx^3+Cx^3.
So that z'=-e^xx^3-3e^xx^2+3Cx^2.
But then when I replace z and z' into z'-\frac{3}{x}z I get that it's worth -e^xx^3 rather than 3x^3e^x. So it seems that I have a "-3" missing factor. I've rechecked all the algebra like 3 times, including now by typing this post and I still don't see where my mistake lies. I'm almost 100% sure it's in the integrating factor method but I really don't see it. I've even reopened Boas' mathematical methods book for the integrating factor method and I feel I've done it right.
Thanks for all help!

 

[Hootenanny]

I would have another look at this bit:

The integrating factor is e^{\int -3 /x dx}=x^{-3}. So that the general solution of this DE (the z's one) should be z=-e^xx^3+Cx^3.

and in particular, the computation of the integrating factor

 

[fluidistic]

I would have another look at this bit:

and in particular, the computation of the integrating factor

I appreciate your help. However I'm afraid I don't see any error for the integrating factor.

 

[Hootenanny]

I appreciate your help. However I'm afraid I don't see any error for the integrating factor.

It's not the IF itself, but rather what you do with it. After multiplication by the IF, your ODE becomes

\frac{d}{dx}\left(\frac{z}{x^3}\right) = 3e^x

Do you see where your factor of three is missing now?

 

[fluidistic]

It's not the IF itself, but rather what you do with it. After multiplication by the IF, your ODE becomes

\frac{d}{dx}\left(\frac{z}{x^3}\right) = 3e^x

Do you see where your factor of three is missing now?

Oh nice, yes now, thank you very very very much.
This works. Continuing on, I reach as final answer y(x)=(3e^xx^3+cx^3)^{-1/3}=\frac{1}{x \sqrt [3] {3e^x+c}}.

 

[Hootenanny]

Oh nice, yes now, thank you very very very much.

My pleasure

This works. Continuing on, I reach as final answer y(x)=(3e^xx^3+cx^3)^{-1/3}=\frac{1}{x \sqrt [3] {3e^x+c}}.

That is indeed correct (or more accurately, the only real solution for appropriate x).