Find the general solution of the differential equation $\mathrm{x}\left({\mathrm{y}}^3+{\mathrm{x}}^3\right)\mathrm{dy}=\left(2{\mathrm{y}}^4+5{\mathrm{x}}^3\mathrm{y}\right)\mathrm{dx}$

Given:$\mathrm{x}\left({\mathrm{y}}^3+{\mathrm{x}}^3\right)\mathrm{dy}=\left(2{\mathrm{y}}^4+5{\mathrm{x}}^3\mathrm{y}\right)\mathrm{dx}$
Simplify the given equation
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{2{\mathrm{y}}^4+5{\mathrm{x}}^3\mathrm{y}}{{\mathrm{xy}}^3+{\mathrm{x}}^4}.........\left(1\right)$
Let, $\mathrm{y}=\mathrm{vx}\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}$
Equation (1) becomes $\mathrm{v}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{2{\mathrm{v}}^4+5\mathrm{v}}{{\mathrm{v}}^3+1}$
$\begin{array}{l}\Rightarrow \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{{\mathrm{v}}^4+4\mathrm{v}}{{\mathrm{v}}^3+1}\\ \Rightarrow \frac{{\mathrm{v}}^3+1}{{\mathrm{v}}^4+4\mathrm{v}}\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}}\\ \Rightarrow \int \frac{4{\mathrm{v}}^3+4}{{\mathrm{v}}^4+4\mathrm{v}}\mathrm{dv}=4\int \frac{\mathrm{dx}}{\mathrm{x}}\\ \Rightarrow \log \left|{\mathrm{v}}^4+4\mathrm{v}\right|=\log (\mathrm{x}{)}^4+\log \mathrm{C}\\ \Rightarrow \log \left|\frac{{\mathrm{y}}^4+4{\mathrm{yx}}^3}{{\mathrm{x}}^4}\right|=\log {\mathrm{Cx}}^4\\ \Rightarrow {\mathrm{y}}^4+4{\mathrm{yx}}^3={\mathrm{Cx}}^8\end{array}$
Therefore, the general solution of the given differential equation is ${\mathrm{y}}^4+4{\mathrm{yx}}^3={\mathrm{Cx}}^8$