Make and solve the given equation d^{3} frac{y}{dx^{3}} - d^{2} frac{y}{dx^{2}}=2x + 3

Given $d^3\frac{y}{d{x}^3}-d^2\frac{y}{d{x}^2}=2x+3$
The above equation can be written as $(D^3-D^2)y=2x+3$
Now the auxiliary equation is $m^3-m^2=0$
which gives $m=0,0,1$
Hence the complimentary function is $C.F.=C_1+C_{2}x+c_3{e}^x$
Now we need to find the Particular Integral
$y_p=\frac{1}{D^2+3D-10}x(x^x+1)=\frac{1}{(D+5)(D-2)}x{e}^x+1$
$=e^x\frac{1}{(D+1+5)(D+1-2)}x+\frac{1}{(D+5)(D-2)}x$
$=e^x\frac{1}{(D+6)(D-1)}x+\frac{1}{(D+5)(D-2)}x$
$=e^x\frac{1}{-6(1+\frac{D}{6})(1-D)}x+\frac{1}{-10(1+\frac{D}{5})(1-\frac{D}{2})}x$
$=-\frac{e^x}{6}{(1+\frac{D}{6})}^{-1}(1-D{)}^{-1}x-\frac{1}{10}{(1-\frac{D}{5})}^{-1}{(1-\frac{D}{2})}^{-1}x$
$=-\frac{e^x}{6}(1+\frac{D}{6}+\cdots )(1+D+D^2)x-\frac{1}{10}(1+\frac{D}{5}\cdots )(1+\frac{D}{2}+\cdots )x$
$=-\frac{e^x}{6}(x+1+\frac{1}{6})-\frac{1}{10}(x+\frac{1}{5}+\frac{1}{2})$
$=-\frac{e^x}{6}(x+\frac{7}{6})-\frac{1}{10}(x+\frac{7}{10})$
Hence the general solution is
$y=y_c+y_p=C_1{e}^{-5}x+C_2{e}^{2}x\pm e_6(x+\frac{7}{6})-\frac{1}{10}(x+\frac{7}{10})$