$ (\frac{d}{dx} + \frac{1}{x} ) (\frac{d}{dx} + \frac{1}{x} ) y = \frac{1}{x^4} \\ $

$ (\frac{d}{dx} + \frac{1}{x} ) (\frac{dy}{dx} + \frac{y}{x} ) = \frac{1}{x^4} \\ $

$ \frac{d^2y}{dx^2} +\frac{d}{dx} (\frac{y}{x}) + \frac{1}{x}\frac{dy}{dx} + \frac{y}{x^2} = \frac{1}{x^4} \\ $

$ \frac{d^2y}{dx^2} +\frac{1}{x} (\frac{dy}{dx}) - \frac{y}{x^2} + \frac{1}{x}\frac{dy}{dx} + \frac{y}{x^2} = \frac{1}{x^4} \\ $

$ \frac{d^2y}{dx^2} + \frac{2}{x}\frac{dy}{dx} = \frac{1}{x^4} \\ $

$ x^2\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} = \frac{1}{x^2} \\ $

$ x= e^z, z = logx \\ $

$ D(D-1)y + 2Dy = e^{-2z} \\ $

$ (D^2 - D + 2D ) y = e^{-2z} \\ $

$ \text{ The Auxiliary equation is } \\ $

$ D^2 + D =0 \\ $

$ D = 0 , -1 \\ $

$ \text{ C.F. is } y_c = c_1 + c_2e^{-z} \\ $

$ \text{P.I = } \frac{1}{D^2 + D} e^{-2z} \\ $

$ = \frac{1}{4-2} e^{-2z} \\ $

$ = \frac{1}{2} e^{-2z} \\ $

$\therefore \text{ The complete solution is y = C.F. + P.I. } \\ $

$ \therefore y = c_1 + c_2e^{-z} + \frac{1}{2} e^{-2z} \\ $

$ \therefore y = c_1 + \frac{c_1}{x} + \frac{1}{2x^2} \\ $