The general solution of the differential equation $\left(x-y^{2}\right) \mathrm{d} x+y\left(5 x+y^{2}\right) \mathrm{d} y=0$ is :

$\left(x-y^{2}\right) d x+y\left(5 x+y^{2}\right) d y=0$ <br/><br/> $y \frac{d y}{d x}=\frac{y^{2}-x}{5 x+y^{2}}$<br/><br/> Let $y^{2}=t$ $\frac{1}{2} \cdot \frac{d t}{d x}=\frac{t-x}{5 x+t}$ <br/><br/> Now substitute, $t=v x$ <br/><br/> $$ \begin{aligned} & \frac{d t}{d x}=v+x \frac{d v}{d x} \\\\ & \frac{1}{2}\left\{v+x \frac{d v}{d x}\right\}=\frac{v-1}{5+v} \\\\ & x \frac{d v}{d x}=\frac{2 v-2}{5+v}-v=\frac{-3 v-v^{2}-2}{5+v} \\\\ & \int \frac{5+v}{v^{2}+3 v+2} d v=\int-\frac{d x}{x} \\\\ & \int \frac{4}{v+1} d v-\int \frac{3}{v+2} d v=-\int \frac{d x}{x} \\\\ & 4 \ln |v+1|-3 \ln |v+2|=-\ln x+\ln C \\\\ & \left|\frac{(v+1)^{4}}{(v+2)^{3}}\right|=\frac{c}{x} \\\\ & \left| \frac{\left(\frac{y^{2}}{x}+1\right)^{4}}{\left(\frac{y^{2}}{x}+2\right)^{3}}\right|=\frac{c}{x} \\\\ & \left|\left(y^{2}+x\right)^{4}\right|=C\left|\left(y^{2}+2 x\right)^{3}\right| \end{aligned} $$