Mathematics - Differential Equation Question with Solution | TestHub

$\begin{array}{rl}& \left(x^2+y^2\right)dx=5xydy\\ & \Rightarrow \frac{dy}{dx}=\frac{x^2+y^2}{5xy}\end{array}$
Put $y=Vx$ $\begin{array}{rl}& \Rightarrow \mathrm{V}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1+{\mathrm{V}}^2}{5\mathrm{V}}\\ & \Rightarrow \frac{\mathrm{xdv}}{\mathrm{dx}}=\frac{1-4{\mathrm{V}}^2}{5\mathrm{V}}\\ & \Rightarrow \int \frac{\mathrm{V}}{1-4{\mathrm{V}}^2}\mathrm{dV}=\int \frac{\mathrm{dx}}{5\mathrm{x}}\end{array}$
Let $1-4{\mathrm{V}}^2=\mathrm{t}$ $\Rightarrow -8\mathrm{V}\mathrm{dV}=\mathrm{dt}$ $\begin{array}{rl}& \Rightarrow \int \frac{\mathrm{dt}}{(-8)(\mathrm{t})}=\int \frac{\mathrm{dx}}{5\mathrm{x}}\\ & \Rightarrow \frac{-1}{8}\ln |\mathrm{t}|=\frac{1}{5}\ln |\mathrm{x}|+\ln \mathrm{C}\\ & \Rightarrow -5\ln |\mathrm{t}|=8\ln |\mathrm{x}|+\ln \mathrm{K}\\ & \Rightarrow \ln {\mathrm{x}}^8+\ln \left|{\mathrm{t}}^5\right|+\ln \mathrm{K}=0\\ & \Rightarrow {\mathrm{x}}^8\left|{\mathrm{t}}^5\right|=\mathrm{C}\\ & \Rightarrow {\mathrm{x}}^8{\left|1-4{\mathrm{V}}^2\right|}^5=\mathrm{C}\\ & \Rightarrow {\mathrm{x}}^8{\left|\frac{{\mathrm{x}}^2-4{\mathrm{y}}^2}{{\mathrm{x}}^2}\right|}^5=\mathrm{C}\\ & \Rightarrow {\left|{\mathrm{x}}^2-4{\mathrm{y}}^2\right|}^5=\mathrm{Cx}{\mathrm{x}}^2\\ & \text{ given }\mathrm{y}(1)=0\\ & \Rightarrow |1{|}^5=\mathrm{C}\Rightarrow \mathrm{C}=1\\ & \Rightarrow {\left|{\mathrm{x}}^2-4{\mathrm{y}}^2\right|}^5={\mathrm{x}}^2\end{array}$