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\(\displaystyle y_p = - {y_1}\int{{\dfrac{{{y_2}6x^3e^2x }}{{W\left( {{y_1},{y_2}} \right)}}\,dx}} + {y_2}\int{{\dfrac{{{y_1}6x^3e^2x }}{{W\left( {{y_1},{y_2}} \right)}}\,dx}} \) ; where \(\displaystyle \; \; W(y_1 , y_2 )=y_1 'y_2 -y_1 y_2 '\).

the general solution is \(\displaystyle y=y_h +y_p\).

\(\displaystyle \frac{d^2 y_h }{dt^2 }-3\frac{dy_h }{dt} +2y_h =0\) ; \(\displaystyle \; y_h =e^{rx} \implies r^2 -3r+2=(r_1 -1)(r_2 -2)=0 \) ; \(\displaystyle \; (r_1 , r_2 ) = (1 , 2 )\).

\(\displaystyle y_h = C_1 e^x +C_2 e^{2x}\) ; \(\displaystyle \; (y_1 , y_2 ) = (C_1 e^x , C_2 e^{2x} )\).

I think you intended $x^2y'' - 2xy' + 2y = 6x^3e^{2x}$.x^2y"-2xy'+2y=6x^3e^2x

If you did, the solution will be $y = \text{C}_1x^2 + \text{C}_2x + \large\frac32xe^{2x}\!$.

Unfortunately, idontknow got careless, to say the least.

By the way, the solution is very easy to find by using $x^{-3}$ as an integrating factor.

Edit

set \(\displaystyle x=e^t \) ; \(\displaystyle \; x^2y_h'' -2xy'_h+2y_h =e^{2t} \frac{d}{de^t} \left( \frac{d}{de^t} y_h \right)-2e^t \frac{dy_h }{de^t} +2y_h =e^t \frac{d}{dt}\left( e^{-t}\frac{dy_h }{dt}\right) -2\frac{dy_h}{dt}+2y_h =e^t [ e^{-t}\frac{d^2 y_h }{dt^2 }-e^{-t}\frac{dy_h }{dt}]-2\frac{dy_h }{dt} +2y_h =0.\)

\(\displaystyle \frac{d^2 y_h }{dt^2 }-3\frac{dy_h }{dt} +2y_h =0\) \(\displaystyle \implies r^2 -3r+2=(r_1 -1)(r_2 -2)=0 \) ; \(\displaystyle \; (r_1 , r_2 ) = (1 , 2 )\).

\(\displaystyle y_h = C_1 e^t +C_2 e^{2t}\) ; \(\displaystyle \; (y_1 , y_2 ) = (C_1 x , C_2 x^2 )\).

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