Show that the differential equation $\left(x^{2}+x y\right) d y=\left(x^{2}+y^{2}\right) d x$ is a homogeneous equation and find its solution.

(N/A) The given differential equation is $\left(x^{2}+xy\right) dy=\left(x^{2}+y^{2}\right) dx$.
This can be written as $\frac{dy}{dx}=\frac{x^{2}+y^{2}}{x^{2}+xy} = F(x, y)$.
For homogeneity,$F(\lambda x, \lambda y) = \frac{(\lambda x)^2 + (\lambda y)^2}{(\lambda x)^2 + (\lambda x)(\lambda y)} = \frac{\lambda^2(x^2+y^2)}{\lambda^2(x^2+xy)} = F(x, y) = \lambda^0 F(x, y)$.
Since the degree is $0$,the equation is homogeneous.
Substitute $y=vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.
Substituting into the equation: $v + x\frac{dv}{dx} = \frac{x^2 + v^2x^2}{x^2 + vx^2} = \frac{1+v^2}{1+v}$.
$x\frac{dv}{dx} = \frac{1+v^2}{1+v} - v = \frac{1+v^2-v-v^2}{1+v} = \frac{1-v}{1+v}$.
Separating variables: $\frac{1+v}{1-v} dv = \frac{dx}{x}$.
$\frac{-(v-1)-2}{v-1} dv = \frac{dx}{x} \Rightarrow (-1 - \frac{2}{v-1}) dv = \frac{dx}{x}$.
Integrating both sides: $-v - 2\ln|v-1| = \ln|x| + C$.
Substituting $v = \frac{y}{x}$: $-\frac{y}{x} - 2\ln|\frac{y}{x}-1| = \ln|x| + C$.
$-\frac{y}{x} - 2\ln|\frac{y-x}{x}| = \ln|x| + C$.
$-\frac{y}{x} - 2\ln|y-x| + 2\ln|x| = \ln|x| + C$.
$-\frac{y}{x} - 2\ln|y-x| + \ln|x| = C$.