If $$\frac{dy}{dx} + e^x(x^2 - 2)y = (x^2 - 2x)(x^2 - 2)e^{2x}$$ and $$y(0) = 0$$, then the value of $$y(2)$$ is

Given,

$$\frac{dy}{dx}+e^x(x^2-2)y=(x^2-2x)(x^2-2)e^{2x}$$

This is a linear differential equation of the form

$$\frac{dy}{dx}+P(x)y=Q(x)$$

where

$$P(x)=e^x(x^2-2)$$

Now,

$$\int P(x)\,dx=\int e^x(x^2-2)\,dx$$

Observe that

$$\frac{d}{dx}\left[e^x(x^2-2x)\right]=e^x(x^2-2)$$

Hence,

$$\int e^x(x^2-2)\,dx=e^x(x^2-2x)$$

Therefore, integrating factor is

$$IF=e^{e^x(x^2-2x)}$$

Multiplying throughout by the integrating factor,

$$\frac{d}{dx}\left(ye^{e^x(x^2-2x)}\right) = (x^2-2x)(x^2-2)e^{2x}\cdot e^{e^x(x^2-2x)} $$

Let

$$t=e^x(x^2-2x)$$

Then,

$$dt=e^x(x^2-2)\,dx$$

So,

$$\int t e^t\,dt=(t-1)e^t+C$$

Hence,

$$ye^{e^x(x^2-2x)} = \left(e^x(x^2-2x)-1\right)e^{e^x(x^2-2x)}+C $$

Therefore,

$$y=e^x(x^2-2x)-1+Ce^{-e^x(x^2-2x)}$$

Using

$$y(0)=0$$

we get

$$0=0-1+C$$

$$C=1$$

Hence,

$$y=e^x(x^2-2x)-1+e^{-e^x(x^2-2x)}$$

Now, at

$$x=2$$

$$y(2)=e^2(4-4)-1+e^0$$

$$=0-1+1=0$$

Hence,

$$\boxed{0}$$