$(5$ pts $)$ Given that $y_1(x)=e^{2x}$ and $y_2(x)=x{e}^{2x}$ are solutions of the associated homogeneous differential equation, find the particular solution to the following differential equation

$y^{\text{'}\text{'}}-4y^{\text{'}}+4y=4e^{2x}lnx,x>0$

a$.y_p(x)=e^{2x}(2x^{2}lnx+3x^2)$

b$.y_p(x)=e^{2x}(2x^{3}lnx+4x^3)$

c$.y_p(x)=e^{2x}(x^{3}lnx-3x^3)$

d$.y_p(x)=e^{2x}(2x^{2}lnx-3x^2)$

e$.y_p(x)=e^{2x}(xlnx-3x)$