Need help solving ordinary differential equation. Thanks!

Question

usukidoll · Answer

$$(3y^2)y'+(y^3)=e^{-x}$$

anonymous · Answer

$$(3y^2)y'+(y^3)=e ^{-x}$$

usukidoll · Answer

you could try separating them sorry I need to finish this long assignment ... last problem FINALLY!

anonymous · Answer

Its actually a bernoulli ODE

anonymous · Answer

I'm just getting stuck on the last few steps

zepdrix · Answer

$$\Large\bf\sf 3y^2 y'+y^3\quad=\quad e^{-x}$$

$$\Large\bf\sf u\quad=\quad y^3,\qquad\qquad\qquad u'\quad=\quad 3y^2 y'$$

Giving us:$$\Large\bf\sf u'+u\quad=\quad e^{-x}$$

zepdrix · Answer

Mmm does that look right so far? I haven't done Bernoulli in a while :P

zepdrix · Answer

Then just find an integrating factor, yes? :o
Oh this thread is 7hours old lol.
You're probably not even here -_-

usukidoll · Answer

sorry I haven't learn Bernoulli

anonymous · Answer

yes that is right. Sorry for taking so long to respond. 
I got to this point, but I ended up getting the wrong answer. :(

perl · Answer

so do you need help further?

anonymous · Answer

yes

anonymous · Answer

When you use integrating factor as suggested before,
you get
$$

 u(x) =e^{-x} + C  e^{-x}
$$

anonymous · Answer

http://www.wolframalpha.com/input/?i=u%27%28x%29+%2B+u%28x%29+%3De^%28-x%29

anonymous · Answer

$$\Large
y= \sqrt[3]{e^{-x} + C  e^{-x}}
$$

anonymous · Answer

http://www.wolframalpha.com/input/?i=%283y^2%29y%27%2B%28y^3%29%3De^%28-x%29

perl · Answer

still need help?

anonymous · Answer

no I got it thanks