91Ó°ÊÓ

The differential equation is $(2x+y)dy−(x−2y)dx=0$.

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$\mathbf{(}\mathit{D}\mathbf{−}\mathit{a}\mathbf{)}\mathbf{(}\mathit{D}\mathbf{−}\mathit{b}\mathbf{)}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{\text{ }}\mathit{a}\mathbf{≠}\mathit{b}$ 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.

$(2x+y)dy−(x−2y)dx=0$

Rearrange above equation.

$(2x+y)dy+(2y−x)dx=0$

The above equation is in the form of$Mdx+Ndy=0$ Where $M=2y−x,N=2x+y$

Differentiate M with respect to y.

$\frac{dM}{dy}=2$

Differentiate N with respect to x.

$\frac{dN}{dx}=2$

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

Thus, the differential equation is an exact differential equation.

The solution of differential equation is,

$\begin{array}{c}∫Mdx+∫Ndy=C\\ ∫(2y−x)dx+∫\left(2\right(0)+y)dy=C\\ ∫(2y−x)dx+∫\left(y\right)dy=C\\ 2yx−\frac{x^2}{2}+\frac{y^2}{2}=C\end{array}$

Solve the equation further and get,

$4xy−x^2+2y^2=2C$

Thus, the solution of given differential equation is $4xy−x^2+2y^2=2C$.