QJEEM24Mad23c: JEE Main 2024 Mathematics - Class 12 JEEM24Mad23c Solution: The solution of the differential equation , is :

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

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

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