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.

${\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{-}\mathit{x}\mathit{y}\mathbf{=}\mathbf{1}\mathbf{/}\mathit{x}$

The differential equation is $x^2{y}^{\text{'}}-xy=\frac{1}{x}$.

Whenf and 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.

$x^2{y}^{\text{'}}-xy=\frac{1}{x}$

Rearrange above equation.

$\begin{array}{c}\frac{x^2{y}^{\text{'}}-xy}{x^2}=\frac{1}{x\times x^2}\\ y^{\text{'}}-\frac{y}{x}=\frac{1}{x^3}\end{array}$

The above equation is in the form of$y^{\text{'}}+Py=Q$ where$P=-\frac{1}{x},Q=\frac{1}{x^3}$.

The integrating factor is,

$\begin{array}{c}I=\int Pdx\\ =\int -\frac{1}{x}dx\\ =-\ln x\end{array}$

The solution of the differential equation is,

$\begin{array}{c}y{e}^I=\int Q{e}^{I}dx\\ y{e}^{-\ln x}=\int \frac{1}{x^3}{e}^{-\ln x}dx\\ y{x}^{-1}=\int \frac{1}{x^3}{x}^{-1}dx\\ y{x}^{-1}=\int \frac{1}{x^4}dx\end{array}$

Further solve,

$\begin{array}{c}y{x}^{-1}=\int \frac{1}{x^4}dxy{x}^{-1}\\ =-\frac{1}{3x^3}+Cy\\ =-\frac{x}{3x^3}+Cxy\\ =-\frac{1}{3x^2}+Cx\end{array}$

Thus, the solution of the given differential equation is$y=-\frac{1}{3x^2}+Cx$.