The characteristic equation for this ODE is
[tex]r^2+8r+15=(r+5)(r+3)=0[/tex]
and has roots at [tex]r=-5,r=-3[/tex], which admits the characteristic solution
[tex]y_c=C_1e^{-5x}+C_2e^{-3x}[/tex]
For the particular solution, we can try finding a quadratic polynomial
[tex]y_p=ax^2+bx+c[/tex]
[tex]{y_p}'=2ax+b[/tex]
[tex]{y_p}''=2a[/tex]
and substituting into the ODE gives
[tex]2a+8(2ax+b)+15(ax^2+bx+c)=4x^2[/tex]
[tex]15ax^2+(16a+15b)x+(2a+8b+15c)=4x^2[/tex]
[tex]\implies\begin{cases}15a=4\\16a+15b=0\\2a+8b+15c=0\end{cases}\implies a=\dfrac4{15},b=-\dfrac{64}{225},c=\dfrac{392}{3375}[/tex]
so that the particular solution is
[tex]y_p=\dfrac4{15}x^2-\dfrac{64}{225}x+\dfrac{392}{3375}[/tex]
and the general solution is
[tex]y=y_c+y_p[/tex]
[tex]y=C_1e^{-5x}+C_2e^{-3x}+\dfrac4{15}x^2-\dfrac{64}{225}x+\dfrac{392}{3375}[/tex]
