Question: Identify each of the differential equations in Problems 1to 24 as to type (for example, separable, linear first order, linear second order, etc.), and then solve it.

$\left(xcosy-e^{-siny}\right)\mathit{d}\mathit{y}\mathbf{+}\mathit{d}\mathit{x}\mathbf{=}\mathbf{0}$

The differential equation is $\left(x\cos y-e^{-\sin y}\right)dy+dx=0$.

When fand its derivatives are inserted into the equation, a solution is a function y = f(x) that solves the differential equation. The highest order of any derivative of the unknown function appearing in the equation is the order of a differential equation.

A differential equation of the form$\mathbf{(}\mathit{D}\mathbf{-}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{-}\mathit{b}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{a}\mathbf{\ne }\mathit{b}$ has general solution$\mathit{y}\mathbf{=}{\mathit{c}}_{\mathbf{1}}{\mathit{e}}^{\mathbf{a}\mathbf{x}}\mathbf{+}{\mathit{c}}_{\mathbf{2}}{\mathit{e}}^{\mathbf{b}\mathbf{x}}\mathbf{.}$

Consider the equation.

$\left(x\cos y-e^{-\sin y}\right)dy+dx=0$

Rearrange above equation.

$\begin{array}{c}\frac{\left(x\cos y-e^{-\sin y}\right)dy}{dy}+\frac{dx}{dy}=0\\ x\cos y-e^{-\sin y}+\frac{dx}{dy}=0\\ \frac{dx}{dy}+x\cos y=e^{-\sin y}\end{array}$

The above equation is in the form of$y^{\text{'}}+Py=Q$ where $P=\cos y,Q=e^{-\sin y}$.

The integrating factor is,

$\begin{array}{c}I=\int Pdy\\ =\int \cos ydy\\ =\sin y\end{array}$

The solution of differential equation is,

$\begin{array}{c}x{e}^I=\int Q{e}^{I}dyx{e}^{\sin y}\\ =\int e^{-\sin y}{e}^{\sin y}dyx{e}^{\sin y}\\ =\int 1dyx{e}^{\sin y}\\ =y+Cx\end{array}$

Solve the equation further and get

$x=\frac{y+C}{e^{\sin y}}$

Thus, the solution of given differential equation is $x=\frac{y+C}{e^{\sin y}}$.