1. Given differential equation

   $x\frac{dy}{dx} + 2y = x^2, \qquad x\neq 0$

   with initial condition

   $y(1)=1.$

2. Put it in standard linear form

   Divide the equation by $x$:

   $\frac{dy}{dx} + \frac{2}{x}y = x.$

   This is a linear differential equation of the form

   $\frac{dy}{dx} + P(x)y = Q(x),$

   where

   $P(x)=\frac{2}{x}, \qquad Q(x)=x.$

3. Find the integrating factor (I.F.)

   $\text{I.F.} = e^{\int P(x)\,dx} = e^{\int \frac{2}{x}\,dx} = e^{2\ln|x|} = x^2$

   (since $x\neq 0$, we can take $x^2$ as the integrating factor).

4. Multiply the differential equation by the integrating factor

   $x^2\frac{dy}{dx} + 2xy = x^3.$

   The left-hand side is:

   $\frac{d}{dx}(x^2y) = x^3.$

5. Integrate both sides

   $x^2y = \int x^3\,dx = \frac{x^4}{4} + C.$

   Therefore,

   $y = \frac{x^4}{4x^2} + \frac{C}{x^2} = \frac{x^2}{4} + \frac{C}{x^2}.$

6. Use the initial condition $y(1)=1$

   Substitute $x=1$ and $y=1$:

   $1 = \frac{1^2}{4} + \frac{C}{1^2} = \frac{1}{4} + C.$

   So,

   $C = 1 - \frac{1}{4} = \frac{3}{4}.$

7. Final solution

   $y = \frac{x^2}{4} + \frac{3}{4x^2}.$

8. Compare with options

   • A: $\frac{4}{5}x^3 + \frac{1}{5x^2}$ — incorrect
   • B: $\frac{3}{4}x^2 + \frac{1}{4x^2}$ — incorrect
   • C: $\frac{x^2}{4} + \frac{3}{4x^2}$ — correct
   • D: $\frac{x^3}{5} + \frac{1}{5x^2}$ — incorrect

Hence, the correct option is C.