Given that, $\text{x}^2\frac{\text{dy}}{\text{dx}}=\text{x}^2+\text{xy}+\text{y}^2$
Dividing both sides by $x^2,$ we get
$\Rightarrow\frac{\text{dy}}{\text{dx}}=1+\frac{\text{y}}{\text{x}}+\frac{\text{y}^2}{\text{x}^2}\ ....(\text{i})$
Let $\text{f}(\text{x, y})=1+\frac{\text{y}}{\text{x}}+\frac{\text{y}^2}{\text{x}^2}$
$\text{f}(\lambda\text{x},\lambda\text{y})=1+\frac{\lambda\text{y}}{\lambda\text{x}}+\frac{\lambda^2\text{y}^2}{\lambda^2\text{x}^2}$
$\text{f}(\lambda\text{x},\lambda\text{y})=\lambda^0\Big(1+\frac{\text{y}}{\text{x}}+\frac{\text{y}^2}{\text{x}^2}\Big)$
$\text{f}(\lambda\text{x},\lambda\text{y})=\lambda^0\text{f}(\text{x},\text{ y})$
Which is homogeneous expression of degree 0.
Put $\text{y}=\text{vx}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\text{v}+\text{x}\frac{\text{dy}}{\text{dx}}$
On substituting these values in Eq. (i), we get
$\Big(\text{v}+\text{x}\frac{\text{dv}}{\text{dx}}\Big)=1+\text{v}+\text{v}^2$
$\Rightarrow\text{x}\frac{\text{dv}}{\text{dx}}=1+\text{v}+\text{v}^2-\text{v}$
$\Rightarrow\text{x}\frac{\text{dv}}{\text{dx}}=1+\text{v}^2$
$\Rightarrow\frac{\text{dv}}{1+\text{v}^2}=\frac{\text{dx}}{\text{x}}$
On integrating both sides, we get
$\int\frac{\text{dv}}{1+\text{v}^2}=\int\frac{\text{dx}}{\text{x}}$
$\tan^{-1}\text{v}=\log|\text{x}|+\text{C}$
$\Rightarrow\tan^{-1}\Big(\frac{\text{y}}{\text{x}}\Big)=\log|\text{x}|+\text{c}$
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