Given that $y_1=x,y_2=x^3$ are linearly independent solutions for the differential equation $y^{\text{'}\text{'}}-\frac{3}{x}{y}^{\text{'}}+\frac{3}{x^2}y=0.$ Using Variation of parameter for $y^{\text{'}\text{'}}-\frac{3}{x}{y}^{\text{'}}+\frac{3}{x^2}y=2x^2{e}^x,$ we get $y_p=$

the above option

$4x^2{e}^x-4x{e}^x$

the above option

$2x^2{e}^x+2x{e}^x$