if you are careful, you can use integration by parts, which is what sort of looks like what you are doing.Thats the farthest I can take you.But I would suggest try doing integration by parts- with remembering that you need to integrate with respect to x.Hope this helps.......
Multiply equation (2) by the integrating factor 1/x^4
dv/dx (1/x^4) - 4v /x5= 8/x^3 -----------------(3)
The left-hand side is the derivative of v /x^4 or (v/x^4) '
(3) becomes
(v/x^4)' = 8/x^3
Integrate both sides
v / x^4 = ∫ 8/x^3 dx + C
v/x^4 = -4/x^2 + C
v = -4x^2+Cx^4
v= Cx^4-4x^2
substitute v=y^2
y^2 = Cx^4-4x^2
y = +/- sqrt [ Cx^4-4x^2]
I gave your question some thought, and here is what I came up with. I don't know if this is the method you NEED to use, but it might be a little easier than what you are doing.
What I did was add xydy to both sides of the equation giving me (and I distributed the 2 on the LHS):
(4x^2 + 2y^2)dx = xy dy
Then I divided both sides by xy to get the dy term by itself.
[(4x^2 + 2y^2)dx] / xy = dy
Then divided everything by dx both sides. This gives us something that looks a little more familiar:
dy/dx = (4x^2 + 2y^2) / xy
Now the major problem is the xy term, but if you think about it, it's the same thing as saying:
(4x^2 + 2y^2) * 1/xy
From there, if you are careful, you can use integration by parts, which is what sort of looks like what you are doing.
That's the farthest I can take you. But I would suggest try doing integration by parts- with remembering that you need to integrate with respect to x.
Hope this helps.
