(a)OR(b)

Find the particular solution of the differential equation given by \[2 \mathrm{x} \mathrm{y}+\mathrm{y}^{2}-2 \mathrm{x}^{2} \frac{d \mathrm{y}}{d \mathrm{x}}=0 ; \mathrm{y}=2, \text { when } \mathrm{x}=1 . \]

Given differential equation can be written as $$ \frac{d \mathrm{y}}{d \mathrm{x}}=\frac{2 \mathrm{x} \mathrm{y}+\mathrm{y}^{2}}{2 \mathrm{x}^{2}}=\frac{\mathrm{y}}{\mathrm{x}}+\frac{\mathrm{y}^{2}}{2 \mathrm{x}^{2}}Let $\displaystyle \mathrm{y}=v \mathrm{x} \Rightarrow \frac{d \mathrm{y}}{d \mathrm{x}}=v+\mathrm{x} \frac{d v}{d \mathrm{x}}$ The equation becomes\begin{aligned} & \mathrm{x} \frac{d v}{d \mathrm{x}}=\frac{1}{2} v^{2} & \Rightarrow \frac{d v}{v^{2}}=\frac{1}{2} \times \frac{d \mathrm{x}}{\mathrm{x}} \end{aligned}Integrating both sides, we get\begin{aligned} & \frac{-1}{v}=\frac{1}{2} \log |\mathrm{x}|+C & \Rightarrow-\frac{\mathrm{x}}{\mathrm{y}}=\frac{1}{2} \log |\mathrm{x}|+C & \mathrm{y}=2, \mathrm{x}=1 \text { gives } C=-\frac{1}{2} \end{aligned}The particular solution is-\frac{\mathrm{x}}{\mathrm{y}}=\frac{1}{2} \log |\mathrm{x}|-\frac{1}{2} \text { or, } \mathrm{y}=\frac{2 \mathrm{x}}{1-\log |\mathrm{x}|}Given differential equation can be written as\frac{d \mathrm{x}}{d \mathrm{y}}-\frac{\mathrm{x}}{\mathrm{y}}=2 \mathrm{y}Integrating Factor $\displaystyle =e^{\int \frac{-1}{\mathrm{y}} d \mathrm{y}}=\frac{1}{\mathrm{y}}$ Solution is $\displaystyle \mathrm{x} \frac{1}{\mathrm{y}}=\int 2 d \mathrm{y}$\Rightarrow \frac{\mathrm{x}}{\mathrm{y}}=2 \mathrm{y}+C$\displaystyle \Rightarrow \mathrm{x}=2 \mathrm{y}^{2}+C \mathrm{y}$