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

$\begin{aligned}
& \left(x^2+y^2\right) d x=5 x y d y \
& \Rightarrow \frac{d y}{d x}=\frac{x^2+y^2}{5 x y}
\end{aligned}$

Put $y=V x$y=Vx $\begin{aligned}
& \Rightarrow \mathrm{V}+\mathrm{x} \frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+\mathrm{V}^2}{5 \mathrm{V}} \
& \Rightarrow \frac{\mathrm{xdv}}{\mathrm{dx}}=\frac{1-4 \mathrm{V}^2}{5 \mathrm{V}} \
& \Rightarrow \int \frac{\mathrm{V}}{1-4 \mathrm{V}^2} \mathrm{dV}=\int \frac{\mathrm{dx}}{5 \mathrm{x}}
\end{aligned}$

Let $1-4 \mathrm{~V}^2=\mathrm{t}$1−4 V2=t $\Rightarrow-8 \mathrm{~V} \mathrm{dV}=\mathrm{dt}$⇒−8 VdV=dt $\displaystyle \begin{aligned} & \Rightarrow \int \frac{\mathrm{dt}}{(-8)(\mathrm{t})}=\int \frac{\mathrm{dx}}{5 \mathrm{x}} \\ & \Rightarrow \frac{-1}{8} \ln |\mathrm{t}|=\frac{1}{5} \ln |\mathrm{x}|+\ln \mathrm{C} \\ & \Rightarrow-5 \ln |\mathrm{t}|=8 \ln |\mathrm{x}|+\ln \mathrm{K} \\ & \Rightarrow \ln \mathrm{x}^8+\ln \left|\mathrm{t}^5\right|+\ln \mathrm{K}=0 \\ & \Rightarrow \mathrm{x}^8\left|\mathrm{t}^5\right|=\mathrm{C} \\ & \Rightarrow \mathrm{x}^8\left|1-4 \mathrm{~V}^2\right|^5=\mathrm{C} \\ & \Rightarrow \mathrm{x}^8\left|\frac{\mathrm{x}^2-4 \mathrm{y}^2}{\mathrm{x}^2}\right|^5=\mathrm{C} \\ & \Rightarrow\left|\mathrm{x}^2-4 \mathrm{y}^2\right|^5=\mathrm{Cx} \mathrm{x}^2 \\ & \text { given } \mathrm{y}(1)=0 \\ & \Rightarrow|1|^5=\mathrm{C} \Rightarrow \mathrm{C}=1 \\ & \Rightarrow\left|\mathrm{x}^2-4 \mathrm{y}^2\right|^5=\mathrm{x}^2\end{aligned}$​⇒∫(−8)(t)dt​=∫5xdx​⇒8−1​ln∣t∣=51​ln∣x∣+lnC⇒−5ln∣t∣=8ln∣x∣+lnK⇒lnx8+ln​t5​+lnK=0⇒x8​t5​=C⇒x8​1−4 V2​5=C⇒x8​x2x2−4y2​​5=C⇒​x2−4y2​5=Cxx2 given y(1)=0⇒∣1∣5=C⇒C=1⇒​x2−4y2​5=x2​