The solution of the differential equation [(x^2+y^2) d x-5 x y ~d y=0, y(1)=0], is :

Question

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

Accepted Answer

$$\begin{aligned}
& \left(x^2+y^2\right) d x-5 x y d y=0 \
& \frac{d y}{d x}=\frac{x^2+y^2}{5 x y} \
& \text { Let } y=v x \
& \frac{d y}{d x}=v+x \frac{d v}{d x} \
& V+x \frac{d v}{d x}=\frac{1+v^2}{5 v} \
& x \frac{d v}{d x}=\frac{1+v^2-5 v^2}{5 v}
\end{aligned}$$

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

$$\begin{aligned}
& \Rightarrow\left|x^2-4 y^2\right|^{5 / 8}=|x|^{\frac{1}{4}} \
& \left|x^2-4 y^2\right|^5=x^2
\end{aligned}$$

Solutions · 1