(a) Find the general solution of the differential equation: ( x y - x^2) d y = y^2 d x

Given differential equation can be written as

$$\frac{d \mathrm{y}}{d \mathrm{x}}=\frac{\mathrm{y}^{$\displaystyle 2$}}{\mathrm{x} \mathrm{y}-\mathrm{x}^{$\displaystyle 2$}}=\frac{\left(\dfrac{\mathrm{y}}{\mathrm{x}}\right)^{$\displaystyle 2$}}{\dfrac{\mathrm{y}}{\mathrm{x}}-$\displaystyle 1$}

$$Separating the variables and integrating

$$\int\left($\displaystyle 1$-\frac{1}{u}\right) d u=\int \frac{d \mathrm{x}}{\mathrm{x}} \Rightarrow u-\log u=\log |\mathrm{x}|+c \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}-\log \frac{\mathrm{y}}{\mathrm{x}}=\log |\mathrm{x}|+c

$$Or Rewriting the given differential equation as:

$$\frac{d \mathrm{y}}{d \mathrm{x}}+\frac{$\displaystyle 2$ \mathrm{x}}{$\displaystyle 1$+\mathrm{x}^{$\displaystyle 2$}} \mathrm{y}=\frac{\sqrt{\mathrm{x}^{$\displaystyle 2$}+$\displaystyle 4$}}{$\displaystyle 1$+\mathrm{x}^{$\displaystyle 2$}}

$$$$ \text { Integrating factor }=\mathrm{e}^{\int \frac{$\displaystyle 2$ \mathrm{x}}{$\displaystyle 1$+\mathrm{x}^{$\displaystyle 2$}} d \mathrm{x}}=\mathrm{e}^{\log \left($\displaystyle 1$+\mathrm{x}^{$\displaystyle 2$}\right)}=$\displaystyle 1$+\mathrm{x}^{$\displaystyle 2$}

$$∴ solution of the differential equation is