The solution of the differential equation $(x^2+y^2) \mathrm{d} x-5 x y \mathrm{~d} y=0, y(1)=0$, is :

<p>$$\begin{aligned} & \left(x^2+y^2\right) d x-5 x y d y=0 \\ & \frac{d y}{d x}=\frac{x^2+y^2}{5 x y} \\ & \text { Let } y=v x \\ & \frac{d y}{d x}=v+x \frac{d v}{d x} \\ & V+x \frac{d v}{d x}=\frac{1+v^2}{5 v} \\ & x \frac{d v}{d x}=\frac{1+v^2-5 v^2}{5 v} \end{aligned}$$</p> <p>$$\begin{aligned} & \frac{1}{8} \int \frac{8 \times 5 v d v}{1-4 v^2}=\int \frac{d x}{x} \\ & \frac{-5}{8} \ln \left|1-4 v^2\right|=\ln |x|+\ln c \\ & \Rightarrow \frac{5}{8} \ln \frac{\left|x^2-4 y^2\right|}{x^2}+\ln |x|=\ln c \\ & \left|\frac{x^2-4 y^2}{x^2}\right|^{5 / 8}|x|=c \\ & \frac{\left|x^2-4 y^2\right|^{5 / 8}}{|x|^{\frac{1}{4}}}=c \because y(1)=0 \Rightarrow c=1 \\ \end{aligned}$$</p> <p>$$\begin{aligned} & \Rightarrow\left|x^2-4 y^2\right|^{5 / 8}=|x|^{\frac{1}{4}} \\ & \left|x^2-4 y^2\right|^5=x^2 \end{aligned}$$</p>