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.

$\mathit{s}\mathit{i}{\mathit{n}}^{\mathbf{2}}\mathit{x}\mathit{d}\mathit{y}\mathbf{+}\left[si{n}^{2}x+(x+y)sin2x\right]\mathit{d}\mathit{x}\mathbf{=}\mathbf{0}$

The differential equation is ${\sin }^{2}xdy+\left[{\sin }^{2}x+(x+y)\sin 2x\right]dx=0$.

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.

${\sin }^{2}xdy+\left[{\sin }^{2}x+(x+y)\sin 2x\right]dx=0$

The above equation is in the form of $Mdx+Ndy=0$Where

$M={\sin }^{2}x+(x+y)\sin 2x,N={\sin }^{2}x$

Differentiate M with respect to y.

$\begin{array}{c}\frac{dM}{dy}=0+0+\sin 2x\\ =\sin 2x\end{array}$

Differentiate N with respect to x.

$\begin{array}{c}\frac{dN}{dx}=2\sin x\cos x\\ =\sin 2x\end{array}$

Since,$\frac{dN}{dx}=\frac{dM}{dy}$

Thus, the differential equation is an exact differential equation.

The solution of differential equation is,

$\begin{array}{c}\int Mdx+\int Ndy=C\\ \int \left[{\sin }^{2}x+(x+y)\sin 2x\right]dx+\int {\sin }^2\left(0\right)dy=C\\ \int \left[{\sin }^{2}x+x\sin 2x+y\sin 2x\right]dx=C\\ \frac{1}{2}x-\frac{\sin 2x}{2}-\frac{x\cos 2x}{2}+\frac{\sin 2x}{2\times 2}-\frac{y\cos 2x}{2}=C\end{array}$

Solve the equation further and get

$x-(x+y)\cos 2x=2C$

Further solve,

$\begin{array}{c}\frac{x}{2}-\frac{x\cos 2x}{2}-\frac{y\cos 2x}{2}=C\\ x-\frac{(1-\cos 2x)}{2}-\frac{y\cos 2x}{2}=C\\ x{\sin }^{2}x-y\cos 2x=2C\\ y=\left(2x{\sin }^{2}x-C\right)\sec 2x\end{array}$

Thus, the solution of given differential equation is $y=\left(2x{\sin }^{2}x-C\right)\sec 2x$.