Answer
The solution of differential equation is given by
$$
tan^{-1}(\frac{y}{x})= ln | x |+ c
$$
where c is an arbitrary constant.
Work Step by Step
$$
y^{'}=\frac{dy}{dx}=\frac{x^{2}+xy+y^{2}}{x^{2}} \quad (*)
$$
Divide each term in the numerator and denominator by the highest power of $x$ , namely $x^{2}$. We obtain
$$
y^{'}=\frac{dy}{dx}=1+ (\frac{y}{x})+(\frac{y}{x})^{2} =f(x,y)
$$
where $f(x,y)$ can be expressed as a function of the ratio $\frac{y}{x}$ only .
Thus eq. (*) is homogeneous .
Now we assume a new dependent variable $ v = \frac{y}{x }$
and we can express
$$
y^{'}=\frac{dy}{dx}= x \frac{dv}{dx}+v = 1+v+v^{2}
$$
$$
x \frac{dv}{dx} = 1+v^{2} \quad (**)
$$
The differential equation (**) is separable and can be written as
$$
\frac{dv}{{1+v^{2}}} = \frac{dx}{x}
$$
integrating the left side with respect to $v$ and the right side with respect to $x$ gives
$$
\int{ \frac{dv}{{1+v^{2}}} }= \int{ \frac{dx}{x}}
$$
we can obtain $v$ implicitly in terms of $ x$ as follows
$$
tan^{-1}(v)= ln | x |+ c
$$
where c is an arbitrary constant.
we find the solution of Eq. (*) by replacing $v$ by $ y/x $ in the previous expression. Hence the solution of eq. (*) is given by
$$
tan^{-1}(\frac{y}{x})= ln | x |+ c
$$
where c is an arbitrary constant.
