Answer
$$y = - \frac{1}{5} + \frac{C}{{{e^{5/x}}}}$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} - \frac{{5y}}{{{x^2}}} = \frac{1}{{{x^2}}} \cr
& {\text{The differential equation has the form }}\frac{{dy}}{{dx}} + P\left( x \right)y = Q\left( x \right) \cr
& {\text{With }}P\left( x \right) = - \frac{5}{{{x^2}}}{\text{, }}Q\left( x \right) = \frac{1}{{{x^2}}} \cr
& {\text{Find the integrating factor }}I\left( x \right) = {e^{\int {P\left( x \right)} dx}} \cr
& I\left( x \right) = {e^{ - 5\int {\frac{1}{{{x^2}}}dx} }} = {e^{5/x}} \cr
& {\text{Multiply the differential equation by the integrating factor}} \cr
& {e^{5/x}}\left( {\frac{{dy}}{{dx}} - \frac{{5y}}{{{x^2}}}} \right) = {e^{5/x}}\left( {\frac{1}{{{x^2}}}} \right) \cr
& {e^{5/x}}\frac{{dy}}{{dx}} - \frac{{5{e^{5/x}}y}}{{{x^2}}} = \frac{{{e^{5/x}}}}{{{x^2}}} \cr
& {\text{Write the left side in the form }}\frac{d}{{dx}}\left[ {I\left( x \right)y} \right] \cr
& \frac{d}{{dx}}\left[ {{e^{5/x}}y} \right] = \frac{{{e^{5/x}}}}{{{x^2}}} \cr
& d\left[ {{e^{5/x}}y} \right] = \frac{{{e^{5/x}}}}{{{x^2}}}dx \cr
& {\text{Integrate both sides}} \cr
& {e^{5/x}}y = - \frac{1}{5}\int {{e^{5/x}}\left( { - \frac{5}{{{x^2}}}} \right)} dx \cr
& {e^{5/x}}y = - \frac{1}{5}{e^{5/x}} + C \cr
& {\text{Solve for }}y \cr
& y = - \frac{1}{5} + \frac{C}{{{e^{5/x}}}} \cr} $$
