The solution to the differential equation $\sin \left(\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}\right)\cos \mathrm{y}=\frac{\mathrm{dy}}{\mathrm{dx}}+\sin \mathrm{ycos}\left(\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}\right)$ is

Given equation is $\sin \left(\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}\right)=\frac{\mathrm{dy}}{\mathrm{dx}}$

$\Rightarrow \mathrm{y}=\mathrm{x}\frac{\mathrm{dy}}{\mathrm{dx}}-{\sin }^{-1}\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)$

Again differentiating we get

$\begin{array}{l}\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{x}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}-\frac{1}{\sqrt{1-{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2}}\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\\ \Rightarrow \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=0\text{ or }\mathrm{x}=\frac{1}{\sqrt{1-{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2}}\\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{c}\text{ or }{\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2=1-\frac{1}{{\mathrm{x}}^2}\end{array}$

using, $\frac{\mathrm{dy}}{\mathrm{dx}}=c \mathrm{in} given equation \text{ we get }\mathrm{y}=\mathrm{cx}-{\sin }^{-1}\mathrm{c}$

Also for particular value of c = 0,y = 0 is also a solution.

Finally using ${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^2=1-\frac{1}{{\mathrm{x}}^2}$ in (i) we get

$\text{ i.e., }\mathrm{y}=\sqrt{{\mathrm{x}}^2-1}-{\sin }^{-1}\left(\frac{\sqrt{{\mathrm{x}}^2-1}}{\mathrm{x}}\right)$