I'm not sure what to make of "yp" and "ye ns", or the given initial conditions, so I'll let you handle that and just focus on the general solution.
[tex]2y''+3y'+y=13x^2[/tex]
The corresponding homogeneous ODE has characteristic equation
[tex]2r^2+3r+1=(2r+1)(r+1)=0[/tex]
with roots at [tex]r=-\dfrac12[/tex] and [tex]r=-1[/tex], so the characteristic solution to the ODE is
[tex]y_c=C_1e^{-x/2}+C_2e^{-x}[/tex]
For the non-homogeneous ODE, assume a solution of the form
[tex]y_p=ax^2+bx+c[/tex]
[tex]\implies{y_p}'=2ax+b[/tex]
[tex]\implies{y_p}''=2a[/tex]
Substituting [tex]y_p[/tex] and its derivatives into the ODE gives
[tex]2(2a)+3(2ax+b)+(ax^2+bx+c)=13x^2[/tex]
[tex]ax^2+(6a+b)x+(4a+3b+c)=13x^2[/tex]
[tex]\implies\begin{cases}a=13\\6a+b=0\\4a+3b+c=0\end{cases}\implies a=13,b=-78,c=182[/tex]
so that the particular solution is
[tex]y_p=13x^2-78x+182[/tex]
and the general solution is
[tex]\boxed{y(x)=C_1e^{-x/2}+C_2e^{-x}+13x^2-78x+182}[/tex]
