Linear first-order differentials

[Nakita]

Hi,
I was just going through my study guide for Calc II, and when I came to linear first-order differentials I encountered a real bump. I don't know what I'm doing wrong, but nearly every one I've worked on hasn't come out right.

Here is one:

x y' + 2y = 1 - x^-1

The book says the answer should be (x² - 2x + C)/2x²


First, I try putting it into standard form, dy/dx + P(x)y = Q(x).
dy/dx + 2y/x = 1/x - 1/x²

So, P(x) = 1/x

V(x) = e^{Intergral of [P(x) dx]}
V(x) = e^ln(x)
V(x) = x

Then I multiply V(x) on each side of the standard form and integrate:

V(x) * (dy/dx + 2y/x ) = V(x) * (1/x - 1/x²)
Integral of [ d/dx (x y) ] = Integral of [ 1 - 1/x ]
x y = x - ln(x) + C
y = (x - ln(x) + C)/x

These two answers are similar, so I feel like I'm on the right track, but I must be missing something that's thwarting the whole operation! Will someone please lend their expertise? I must know this!

[BigGlenntheHeavy]

[Nakita]

Thank you, BigGlenn! I see where I went wrong; I knew it had to have been in the beginning. Thanks for responding so quickly, you really helped me!