Could someone check if I did this correctly? Homogeneous Differential Equation 
$$y' = \frac{-y-x}{x}, y' = \frac{-y}{x} -1$$
$$F(v) = -v -1$$ Since $$y = xv$$ then $$y' = v + xv'$$
Therefore:
$$v+xv' = -v-1$$
$$xv' = -2v-1$$
$$\frac{-dv}{2v+1} = \frac{dx}{x}$$
$$-\frac{1}{2}ln\bigg |1+2v\bigg|=ln|x| + C$$
$$ln|1+2v| = -2ln|x| -2C$$
$$1+2v = \pm (x^{-2}*e^{-2C})$$
So here is where I get anxious, I'm not sure what to do with with the constants and the plus/minus from the LHS absolute value sign. But here's what I did and what we learned in class:
$$1+2v = \frac{D}{x^{2}}$$ where $D = \pm e^{-2C}$ and so $v =\frac{D}{2x^{2}} - \frac{1}{2}$
Do I substitute $v = \frac{y}{x}$ back in? So $ \frac{y}{x}= \frac{D}{2x^{2}} - \frac{1}{2}$
$$y = \frac{xD}{2x^{2}} - \frac{x}{2}$$
$$y = \frac{D}{2x} - \frac{x}{2}$$
Plugging in the initial condition where y(1) = 1:
$$1 = \frac{D}{2} - \frac{1}{2}$$
$$D = 3$$
Therefore, the general solution is...?
$$y(x) = \frac{3}{2x} - \frac{x}{2}$$
The thing I'm confused about is, are we solving for $D$ since we substituted $D$ for $e^{-2C}$?
Or does the constant variable doesn't matter?

$$y'x+y+x=0$$
$$(xy)'+x=0$$
Simply integrate 
$$xy =-\int xdx=-\frac {x^2} 2 +K$$
$$y =-\frac {x} 2 +\frac Kx$$
then since $y(1)=1 \to K= \frac 32$
$$y =-\frac {x} 2 +\frac {3}{2x}$$
So your answer is correct..
