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
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.