Bernoulli Equations 
I have this equation I am trying to solve, I'm not sure if I did it correctly, but here's my work:
$$y' +y(x^2-1) + xy^6 =0$$
$$y' + y(x^2-1) = -xy^6 $$
$$v = y^{-5}, v' = -5y^{-6}y'$$
$$y' = \frac{-y^{6}v'}{5}$$
Substituting in:
$$\frac{-y^{6}v'}{5} + y(x^2-1) =-xy^6$$
$$v' -5v(x^2-1) = 5x$$
So this is now the linear form, therefore $$r(x) = e^{-5\int x^2-1} = e^{-5(\frac{x^3}{3}-x)}$$
At this point, I multiplied each side by r(x):
$$e^{-5(\frac{x^3}{3}-x)}v' -5v(x^2-1)e^{-5(\frac{x^3}{3}-x)} = e^{-5(\frac{x^3}{3}-x)}5x$$
And then I rewrote it like this:
$$\frac{d}{dx} \bigg [e^{-5(\frac{x^3}{3}-x)}v)\bigg ] = e^{-5(\frac{x^3}{3}-x)}5x$$
From here it's obvious to integrate both sides, doing so yields:
$$e^{-5(\frac{x^3}{3}-x)}v(x) = 5\int xe^{-5(\frac{x^3}{3}-x)} dx$$
I don't know what to do from here
Thanks
 A: Starting from your second-to-last line
$$ \frac{d}{dx}\left[\exp\left(-\frac{5x^3}{3}+5x\right)v(x) \right] = 5x \exp\left(-\frac{5x^3}{3}+5x\right) $$
A convenient way to proceed is to integrate from the given initial point
$$ \exp\left(-\frac{5x^3}{3}+5x\right)v(x) = \int_1^x 5t \exp\left(-\frac{5t^3}{3}+5t\right)dt + C $$
Plugging in $x=1$ gives $C = e^{10/3}$, so
$$ v(x) = \exp\left(\frac{5x^3}{3}-5x\right)\left[\int_1^x 5t \exp\left(-\frac{5t^3}{3}+5t\right)dt + e^{10/3}\right] $$
Then just take $y = v^{-1/5}$ and you're done
A: I assume your solution up to this point is correct, so we are dealing with:
$$e^{-5(\frac{x^3}{3}-x)}v(x) = 5\int xe^{-5(\frac{x^3}{3}-x)} dx$$
You ask: "what do I do with the initial condition?"
It's $y(1)=1$, which means that $v(1)=y(1)^{-5}=1$.
When you have a solution in the form of an integral, the initial condition determines its limits:
$$e^{-5(\frac{1}{3}-1)}v(1) = 5\int_{x_0}^1 xe^{-5(\frac{x^3}{3}-x)} dx$$
Rewriting:
$$\int_{x_0}^1 xe^{-5(\frac{x^3}{3}-x)} dx=\frac{1}{5} e^{10/3}$$
This equation determines $x_0$. Which means, your solution should be:
$$v(x) = 5 e^{5(\frac{x^3}{3}-x)}\int_{x_0}^x xe^{-5(\frac{x^3}{3}-x)} dx$$
Where $x_0$ is now known.
