The general solution of the differential equation x - y 2 d x + y 5 x + y 2 d y = 0 is

Given, $\left(x-y^2\right)dx+y\left(5x+y^2\right)dy=0$

After rearranging we get,

$\frac{dy}{dx}=\frac{y^2-x}{y\left(5x+y^2\right)}$ 

Now, let $y^2=v\Rightarrow 2y\frac{dy}{dx}=\frac{dv}{dx}$

$\frac{1}{2y}\frac{dv}{dx}=\frac{v-x}{y\left(5x+v\right)}$

$\frac{dv}{dx}=2\left(\frac{v-x}{5x+v}\right)$

Now let $v=kx$

$\Rightarrow k+x\frac{dk}{dx}=2\left(\frac{kx-x}{5x+kx}\right)$

$\Rightarrow x\frac{dk}{dx}=-\frac{\left(k^2+3k+2\right)}{k+5}$

$\Rightarrow \int \frac{\left(5+k\right)}{\left(k+1\right)\left(k+2\right)}dk=\int -\frac{dx}{x}$

$\Rightarrow \int \left(\frac{4}{k+1}-\frac{3}{k+2}\right)dk=-\int \frac{dx}{x}$

$\Rightarrow 4\ln \left(k+1\right)-3\ln \left(k+2\right)=-\ln x+\ln c$

$\Rightarrow \frac{{\left(k+1\right)}^4}{{\left(k+2\right)}^3}=\frac{c}{x}$

$\Rightarrow \frac{{\left({\displaystyle \frac{v}{x}}+1\right)}^4}{{\left({\displaystyle \frac{v}{x}}+2\right)}^3}=\frac{c}{x}$

$\Rightarrow c{\left(y^2+2x\right)}^3={\left(y^2+x\right)}^4$