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[sup:2vaq2uut]-1[/sup:2vaq2uut]

The book says the answer should be (x[sup:2vaq2uut]2[/sup:2vaq2uut] - 2x + C)/2x[sup:2vaq2uut]2[/sup:2vaq2uut]


First, I try putting it into standard form, dy/dx + P(x)y = Q(x).
dy/dx + 2y/x = 1/x - 1/x[sup:2vaq2uut]2[/sup:2vaq2uut]

So, P(x) = 1/x

V(x) = e[sup:2vaq2uut]Intergral of [P(x) dx][/sup:2vaq2uut]
V(x) = e[sup:2vaq2uut]ln(x)[/sup:2vaq2uut]
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/x[sup:2vaq2uut]2[/sup:2vaq2uut])
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]

\(\displaystyle xy'+2y \ = \ 1-1/x, \ \frac{dy}{dx}+\frac{2}{x}(y) \ = \ \frac{x-1}{x^{2}}\)

\(\displaystyle P(x) \ = \ \frac{2}{x}, \ \int P(x)dx \ = \ ln(x^{2}), \ Q(x) \ = \ \frac{x-1}{x^{2}}\)

\(\displaystyle e^{\int P(x)dx} \ = \ x^{2}\)

\(\displaystyle Ergo, \ yx^{2} \ = \ \int\frac{x-1}{x^{2}}(x^{2})dx+C_1\)

\(\displaystyle yx^{2} \ = \ \int(x-1)dx+C_1 \ = \ \frac{x^{2}}{2}-x+C_1\)

\(\displaystyle Hence, \ y \ = \ \frac{1}{2}-\frac{1}{x}+\frac{C_1}{x^{2}} \ = \ \frac{x^{2}-2x+2C_1}{2x^{2}}\)

\(\displaystyle Let \ C \ = \ 2C_1, \ then \ y \ = \ \frac{x^{2}-2x+C}{2x^{2}}\)

 

[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!