Using the method of undetermined coefficients, find the general solution of the following differential equation...

We find the general solution of eq (1) of the method of undesermened coefficient.
Now the aurilary eq for the homogenous eq
${\displaystyle y3y^{\prime }+2y=0}$ is
${\displaystyle m^2-3m+2=0}$
${\displaystyle (m-2)(m-1)=0}$
${\displaystyle m=1,2}$
So the function ${\displaystyle y=c_{1}e+c_2{e}^{2x}}$
Here we take ${\displaystyle y_p=A{x}^2+Bx+c}$,
${\displaystyle y_p=2a+b}$
${\displaystyle y_{p}2a}$
Putting the value of ${\displaystyle y_p,y_p,y_p}$ in eq (1) we get
${\displaystyle 2a-3(2ax+b)+2(a{x}^2+bx+c)=x^2}$
${\displaystyle 2a-6ax-3b+2a{x}^2+2bx+2=x^2}$
${\displaystyle 2a{x}^2+(-6a+2b)x+(2a-3b+2c)=x^2}$
${\displaystyle 2a=1\Rightarrow a=\frac{1}{2}}$
${\displaystyle -6A+2b=0\Rightarrow -6\times \frac{1}{2}+2b=0\Rightarrow -3+2b=0\Rightarrow b=\frac{3}{2}}$
${\displaystyle 2a-3b+2c=0\Rightarrow 2\cdot \frac{1}{2}-3\cdot \frac{3}{2}+2=0}$
${\displaystyle \Rightarrow 1-\frac{9}{2}+2c=0}$
${\displaystyle \Rightarrow 2c\cdot \frac{9}{2}-1}$
${\displaystyle 2c=\frac{1}{2}}$
${\displaystyle =1c=\frac{1}{4}}$
${\displaystyle y_p=\frac{1}{2}{x}^2+\frac{3}{2}x+\frac{1}{4}}$
The general solution is
${\displaystyle y\left(x\right)=y_c+y_p}$
${\displaystyle =c_1{e}^x+c_2{e}^{2x}+\frac{1}{2}{x}^2+\frac{3}{2}x+\frac{1}{4}}$

Given differential equation is
${\displaystyle y3y^{\prime }+2y=x^2+x+1}$ (1)
${\displaystyle AEis{D}^2-3D+2=0}$
${\displaystyle (D-1)(D-2)=0}$
${\displaystyle \therefore D=1,2}$
${\displaystyle \therefore CF=c_1{e}^x+c_2{e}^{2x}}$
PI is of the form
${\displaystyle y=A_0+A_{1}x+A_2{x}^2}$
${\displaystyle y^{\prime }=0+A_1+2A_{2}x}$
${\displaystyle y0+2A_2}$
Substituting in (1)
${\displaystyle 2A_2-3(A_1+2A_{2}x)+2(A_0+A_{1}x+A_2{x}^2)=x^2+x+1}$
${\displaystyle \therefore 2A_2{x}^2+(2A_1-6A_2)x+2A_0-3A_1+2A_2=x^2+x+1}$
Comparing the coefficients
${\displaystyle 2A_2=1,2A_1-6A_2=1,2A_0-3A_1+2A_2=1}$
Solving, ${\displaystyle A_2=\frac{1}{2},A_1=2,A_0=3}$
${\displaystyle \therefore PI=3+2x+\frac{1}{2}{x}^2}$
${\displaystyle \therefore GS=CF+PI}$
${\displaystyle y=c_1{e}^x+c_2{e}^{2x}+(3+2x+\frac{1}{2}{x}^2)}$

Solve $y^{″}-3y^{\prime }+2y=x^2$
Homogen solution: $y=A{e}^x+B{e}^{2x}$
Particular solution:
$y_p=x(A{x}^2+Bx+C)=A{x}^3+B{x}^2+Cx$
$y^{\prime }p=3A{x}^2+2Bx+C$
$y^{″}p=6Ax+2B$
Put his into the initial equartion to get A, B and C gives me:
$A=0,B=1/2,C=3/2$
This leads me to the answer:
$y=A{e}^x+B{e}^{2x}+x^2/2+3x/2$
However the correct answer is
$y=A{e}^x+B{e}^{2x}+x^2/2+3x/2+7/4$
Where's my miss? Where comes the last term from?