Find the general solution of the differential equation y - y^(4) = (y^(3) + 3x)y'

Question

anonymous · Answer

I got an integrating factor of 1/(y^4)

anonymous · Answer

I got the answer to be -y^4-y^4+y-3xy=0

anonymous · Answer

what was your integrating factor?

anonymous · Answer

Hold up, I'm getting too many questions.

terenzreignz · Answer

@xjakester that was a y' on the right-side

anonymous · Answer

i got an answer of C = x(y^(-3)) -x - ln(y), with ln(y) being the constant i found

anonymous · Answer

it is incorrect though

anonymous · Answer

i believe the integrating factor is correct though, since the partial deriv. with respect to y of  (y^(-3) -1) and the partial deriv with respect to x of (-y^(-1) - 3xy^(-4)) are equal to one another

anonymous · Answer

Are you looking for an Alternative form, or solutions?

anonymous · Answer

a general solution to an exact differential quation

terenzreignz · Answer

It's exact? 
Hang on...
$$\Large y - y^4 = (y^3+3x)\color{green}{y'}$$

anonymous · Answer

Solving for y or x?

terenzreignz · Answer

$$\Large y - y^4 = (y^3+3x)\frac{\color{blue}{dy}}{\color{red}{dx}}$$

anonymous · Answer

its not exact initially, it is exact after you multiply it by the integrating factor y^(-4)

terenzreignz · Answer

@xjakester not really... it's calculus ^_^

anonymous · Answer

I skipped calculus for statistic. x_x

terenzreignz · Answer

I see... allow me to prod it a bit ... see if I can get something...
$$\Large (y-y^4)\color{red}{dx}-(y^3+3x)\color{blue}{dy}=0$$

anonymous · Answer

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