Solve the differential equation $\displaystyle \left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) d \mathrm{x}+\mathrm{x} \mathrm{y} d \mathrm{y}=0, \mathrm{y}(1)=1$.

\[\frac{d \mathrm{y}}{d \mathrm{x}}=-\frac{\mathrm{x}^{2}+\mathrm{y}^{2}}{\mathrm{x} \mathrm{y}}=-\frac{1+\left(\dfrac{\mathrm{y}}{\mathrm{x}}\right)^{2}}{\dfrac{\mathrm{y}}{\mathrm{x}}}, \text { Put } \mathrm{y}=v \mathrm{x}, \frac{d \mathrm{y}}{d \mathrm{x}}=v+\mathrm{x} \frac{d v}{d \mathrm{x}} \] For $\displaystyle \mathrm{x}=1, \mathrm{y}=1, \mathrm{D}=3$, ∴ The solution of the differential equation is, $\displaystyle \left(2 \mathrm{y}^{2}+\mathrm{x}^{2}\right) \mathrm{x}^{2}=3$