91Ó°ÊÓ

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

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.

$(y+2x)dx−xdy=0$

The above equation is first order homogeneous differential equation.

Rearrange above equation.

$\begin{array}{c}(y+2x)−x\frac{dy}{dx}=0\\ \frac{(y+2x)}{x}−\frac{x\frac{dy}{dx}}{x}=0\\ \frac{y}{x}+2−\frac{dy}{dx}=0\\ \frac{dy}{dx}=\frac{y}{x}+2\end{array}$

Assume the following.

$\begin{array}{c}v=\frac{y}{x}\\ \frac{dy}{dx}=v+x\frac{dv}{dx}\end{array}$

Substitute above values in above equation.

$\begin{array}{c}v+x\frac{dv}{dx}=v+2\\ x\frac{dv}{dx}=2\\ ∫dv=∫2\frac{dx}{x}\\ v=2\ln x+C\end{array}$

Substitute$\frac{y}{x}$ for v in above equation.

$\begin{array}{l}\frac{y}{x}=2\ln x+C\\ y=2x\ln x+Cx\end{array}$

Thus, the solution of given differential equation is $y=2x\ln x+Cx$.