91Ó°ÊÓ

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

$x^2{y}^{\text{'}}−xy=\frac{1}{x}$

Rearrange above equation.

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

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,

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

The solution of the differential equation is,

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

Further solve,

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

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