Show that the differential equation $y^{\prime} = \frac{x+y}{x}$ is a homogeneous equation and find its general solution.

(N/A) The given differential equation is:
$y^{\prime} = \frac{x+y}{x}$
$\Rightarrow \frac{dy}{dx} = \frac{x+y}{x} \quad \dots (1)$
Let $F(x, y) = \frac{x+y}{x}$.
Now,$F(\lambda x, \lambda y) = \frac{\lambda x + \lambda y}{\lambda x} = \frac{\lambda(x+y)}{\lambda x} = \frac{x+y}{x} = \lambda^0 F(x, y)$.
Since $F(\lambda x, \lambda y) = \lambda^0 F(x, y)$,the given equation is a homogeneous differential equation.
To solve it,substitute $y = vx$.
Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.
Substituting these into equation $(1)$:
$v + x \frac{dv}{dx} = \frac{x + vx}{x} = \frac{x(1+v)}{x} = 1+v$.
$x \frac{dv}{dx} = 1$.
$dv = \frac{dx}{x}$.
Integrating both sides:
$\int dv = \int \frac{dx}{x}$.
$v = \log|x| + C$.
Since $v = \frac{y}{x}$,we have $\frac{y}{x} = \log|x| + C$.
Therefore,the general solution is $y = x \log|x| + Cx$.