$$y' = \frac{-y-x}{x}, y' = \frac{-y}{x} -1$$ $$F(v) = -v -1$$ 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?
Your result is correct. Your final result does not represent the general solution anymore, since you have included the initial conditions, so the solution you got is a specific solution to this differential equation, with the initial conditions you have specified. There is no more constant to solve for, since you have done that already.