Solve the differential equation $\displaystyle (\mathrm{x}-\sin \mathrm{y}) d \mathrm{y}+(\tan \mathrm{y}) d \mathrm{x}=0$, given $\displaystyle \mathrm{y}(0)=0$.

The differential equation can be written as: \[\frac{d \mathrm{x}}{d \mathrm{y}}+\cot \mathrm{y} \cdot \mathrm{x}=\cos \mathrm{y} \text {, which is a linear order differential equation } \] Here, \(\displaystyle \mathbf{P}=\cot \mathbf{y}, \mathbf{Q}=\cos \mathbf{y}\), I.F. (Integrating Factor) \(\displaystyle =\mathbf{e}^{\int \cot \mathbf{y} \mathbf{d y}}=\mathbf{e}^{\log \sin \mathbf{y}}=\sin \mathbf{y}\) The solution is,\(\displaystyle \mathrm{x}(\sin \mathrm{y})=\int \cos \mathrm{y} \cdot \sin \mathrm{y} d \mathrm{y}\) \[\Rightarrow \mathrm{x}(\sin \mathrm{y})=\frac{(\sin \mathrm{y})^{2}}{2}+C, \text { For } \mathrm{x}=0, \mathrm{y}=0, C=0 . \] ∴ The Particular solution is: \(\displaystyle \mathrm{x} \sin \mathrm{y}=\frac{\sin ^{2} \mathrm{y}}{2}\) or \(\displaystyle \sin \mathrm{y}=2 \mathrm{x}\) or \(\displaystyle \mathrm{y}=\sin ^{-1} 2 \mathrm{x}\)