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

Question

Aicharya · Accepted Answer

$\begin{aligned} & \left(x^2+y^2ight) d x=5 x y d y \ & \Rightarrow \frac{d y}{d x}=\frac{x^2+y^2}{5 x y} \end{aligned}$ Put $y=V x$ $\begin{aligned} & \Rightarrow \mathrm{V}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+\mathrm{V}^2}{5 \mathrm{~V}} \ & \Rightarrow \frac{\mathrm{xdv}}{\mathrm{dx}}=\frac{1-4 \mathrm{~V}^2}{5 \mathrm{~V}} \ & \Rightarrow \int \frac{\mathrm{V}}{1-4 \mathrm{~V}^2} \mathrm{dV}=\int \frac{\mathrm{dx}}{5 \mathrm{x}} \end{aligned}$ Let $1-4 \mathrm{~V}^2=\mathrm{t}$ $\Rightarrow-8 \mathrm{~V} \mathrm{dV}=\mathrm{dt}$ $\begin{aligned} & \Rightarrow \int \frac{\mathrm{dt}}{(-8)(\mathrm{t})}=\int \frac{\mathrm{dx}}{5 \mathrm{x}} \ & \Rightarrow \frac{-1}{8} \ln |\mathrm{t}|=\frac{1}{5} \ln |\mathrm{x}|+\ln \mathrm{C} \ & \Rightarrow-5 \ln |\mathrm{t}|=8 \ln |\mathrm{x}|+\ln \mathrm{K} \ & \Rightarrow \ln \mathrm{x}^8+\ln \left|\mathrm{t}^5ight|+\ln \mathrm{K}=0 \ & \Rightarrow \mathrm{x}^8\left|\mathrm{t}^5ight|=\mathrm{C} \ & \Rightarrow \mathrm{x}^8\left|1-4 \mathrm{~V}^2ight|^5=\mathrm{C} \ & \Rightarrow \mathrm{x}^8\left|\frac{\mathrm{x}^2-4 \mathrm{y}^2}{\mathrm{x}^2}ight|^5=\mathrm{C} \ & \Rightarrow\left|\mathrm{x}^2-4 \mathrm{y}^2ight|^5=\mathrm{Cx} \mathrm{x}^2 \ & 	ext { given } \mathrm{y}(1)=0 \ & \Rightarrow|1|^5=\mathrm{C} \Rightarrow \mathrm{C}=1 \ & \Rightarrow\left|\mathrm{x}^2-4 \mathrm{y}^2ight|^5=\mathrm{x}^2\end{aligned}$

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

Solution