x\(\frac{dy}{dx}+\frac{2y}x=\frac1{x^3}\)
⇒ \(\frac{dy}{dx}+\frac{2}{x^2}y=\frac1{x^4}\) which is linear differential equation.
∴ I.F. = \(e^{\int\frac2{x^2}dx} = e^{\frac{-2}x}\)
y x I.F. = \(\int\frac1{x^4}(I.F)dx\)
⇒ y. e\(\frac{-2}x\) = \(\frac{e^{\frac2x}}{x^4}dx\)
Let \(\frac{-2}x=t\)
⇒ \(\frac2{x^2}dx=dt\)
∴ ye\(\frac{-2}x\) = \(\frac12\int\frac{t^2}4e^tdt\)
= \(\frac18\)(t2et - 2tet + et) + c
= \(\frac18\)(t2 - 2t + 1)et + c
left part
y.e-2/x = \(\frac18\)(t - 1)2et + c
⇒ ye-2/x = \(\frac18\) (-2/x - 1)2e-2/x + c ( ∵ t = -2/x)
⇒ y = \(\frac18\)(2/x + 1)2 + ce2/x which is general solution of given differential equation.