If the solution of the differential equation $\frac{dy}{dx}+e^x\left(x^2-2\right)y=\left(x^2-2x\right)\left(x^2-2\right)e^{2x}$ satisfies y(0)=0, then the value of y(2) is ________.

I.F. $=e^{\int e^x\left(x^2-2\right)dx}=e^{\int e^x\left(x^2-2x+2x-2\right)}dx$

$$\begin{gathered} =e^{e^x\left(x^2-2x\right)} \\ y·e^{e^x\left(x^2-2x\right)}=\int e^{e^x\left(x^2-2x\right)}{e}^x\left(x^2-2x\right)\left(x^2-2\right)e^{x}dx \\ Let e^x\left(x^2-2x\right)=t \\ So, y·e^{e^x\left(x^2-2x\right)}=\int e^t·t dt \\ At x =0, t =0 \\ x =2, t =0 \\ =t·e^t-e^t+c \\ x=0;0·1=0-1+c\Rightarrow c=1 \\ for x=2;y·1=0-1+1=0 \\ y\left(2\right)=0 \end{gathered}$$