Given differential equation is \(x^2\frac {d^2y}{dx^2} + x\frac {dy}{dx} -y = \frac {x^3}{1+x^2}\)

Then \(x \frac {dy}{dx} = \frac {dy}{dz} = dy\), where D = \(\frac {y}{dz}\)

\(x^2 \frac {d^2y}{dx^2} \) = D (D-1) y.

Then given differential equation converts to 

D(D-1)y + Dy-y = \(\frac {e^{3z}}{1+e^{2z}}\)

= (D² - D + D-1)y = \(\frac {e^{3z}}{1+e^{2z}}\)

= (D² -1)y = \(\frac {e^{3z}}{1+e^{2z}}\)

It's auxiliarly equation is m²-1 = 0

= m = ± 1

∴ C.F = C₁ e^-z+ C₂ e^z

= \(\frac {C_1}{x} + C_2 x\) (∵e^z = x)

Let y₁ = \(\frac 1x\) & y₂ = x

∴ W(y₁, y₂) = \(\begin {vmatrix}y_1&y_2\\y_1^1&y_2^1\end {vmatrix} = \begin {vmatrix}\frac 1x&x\\\frac {-1}{x^2}&1\end {vmatrix}\)

= \(\frac1x + x . \frac {1}{x^2} = \frac 1x + \frac 1x = \frac 2x\)

∴ y₁ =\(\int\) \(\frac {\begin {vmatrix}0&x\\\frac {x^3}{1+x^2}&1\end{vmatrix}}{W(y_1,y_2)}dx\)

=  \(\int \frac {-\frac {x^4}{1+x^2}}{\frac 2x}dx\)

= \(\frac {-1}{2}\int \frac {x^5}{1+x^2}dx = \frac {-1}{2} \int (x^3-x+\frac {x}{1+x^2})dx\)

= \(\frac {-1}{2}(\frac {x^4}{4} - \frac {x^2}{2} + 12 log (1 +x^x2))\) & 

y₂ = \(\int \frac {\begin {vmatrix}\frac 1x&0\\\frac {-1}{x^2} & \frac {x^3}{1+x^2}\end {vmatrix}}{W (y_1,y_2)}\)

= \(\int \frac {\frac {x^2}{1+x^2}}{\frac 2x}dx = \frac 12\int\frac {x^3}{1+x^2}dx\)

= \(\frac 12 \int (x- \frac {x}{1+x^2})dx\)

 = \(\frac 12(\frac {x^2}{2} - \frac 12 log (1+x^2))\)

∴ P.I = y₁y₁ + y₂y₂

= \(\frac {-1}{2x} (\frac {x^4}{4} - \frac {x^2}{2} + \frac 12 log (1+x^2)) + \frac x2 (\frac {x^2}{2} - \frac 12 log (1+x^2))\)

= \(\frac {-x^3}{8} + \frac x4 - \frac {1}{4x}log (1+x^2) + \frac {x^3}{4} - \frac x4 log (1+x^2)\)

= \(\frac {x^3}{8} + \frac x4 - \frac x4log (1+x^2) \frac {-1}{4x}log (1+x^2)\)

∴ Complete solution is 

y = C.F + P.I

= \(\frac {C_1}{x} + C_2x + \frac {x^3}{8} + \frac x4 - \frac x4 log (1+x^2) - \frac {1}{4x} log (1+x^2)\)