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).