1. Given differential equation

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

with initial condition

$$y(0)=0.$$

This is a linear differential equation of the form

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

where

$$P(x)=e^x(x^2-2), \qquad Q(x)=(x^2-2x)(x^2-2)e^{2x}.$$

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2. Look for a useful substitution

Notice that the right-hand side contains $e^{2x}(x^2-2)$, and the coefficient of $y$ contains $e^x(x^2-2)$. This suggests trying

$$y=v e^x.$$

Then

$$\frac{dy}{dx}=e^x\frac{dv}{dx}+ve^x.$$

Substitute into the differential equation:

$$e^x\frac{dv}{dx}+ve^x+e^x(x^2-2)(ve^x)=(x^2-2x)(x^2-2)e^{2x}.$$

Divide throughout by $e^x$:

$$\frac{dv}{dx}+v+v(x^2-2)e^x=(x^2-2x)(x^2-2)e^x.$$

This does not simplify nicely. So instead, we solve by integrating factor.

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3. Find the integrating factor

For

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

integrating factor is

$$\text{I.F.}=e^{\int P(x)dx}=e^{\int e^x(x^2-2)dx}.$$

Now compute

$$\int e^x(x^2-2)dx.$$

We use the known result:

$$\frac{d}{dx}\big[e^x(x^2-2x)\big]=e^x(x^2-2x)+e^x(2x-2)=e^x(x^2-2).$$

Hence,

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

So the integrating factor is

$$\text{I.F.}=e^{e^x(x^2-2x)}.$$

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4. Write the solution form

Multiplying the differential equation by the integrating factor,

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

That is,

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

Now observe:

Let

$$u=e^x(x^2-2x).$$

Then

$$\frac{d\nu}{dx}=e^x(x^2-2).$$

So the right-hand side becomes

$$(x^2-2x)e^x\cdot e^{\nu}\cdot \frac{d\nu}{dx}.$$

But since

$$\nu=e^x(x^2-2x),$$

we have

$$(x^2-2x)e^x=\nu.$$

Therefore,

$$\frac{d}{dx}\left[y e^{\nu}\right]=\nu e^{\nu}\frac{d\nu}{dx}.$$

Hence,

$$ye^{\nu}=\int \nu e^{\nu}\,d\nu.$$

Now,

$$\int \nu e^{\nu}d\nu=e^{\nu}(\nu-1)+C.$$

Thus,

$$ye^{\nu}=e^{\nu}(\nu-1)+C$$

or

$$y=\nu-1+Ce^{-\nu}.$$

Substituting back,

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

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5. Use the initial condition $y(0)=0$

At $x=0$,

$$e^0(0^2-2\cdot 0)=0.$$

So

$$y(0)=0-1+Ce^0=-1+C.$$

Given $y(0)=0$,

$$-1+C=0 \implies C=1.$$

Hence,

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

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6. Find $y(2)$

At $x=2$,

$$x^2-2x=4-4=0.$$

Therefore,

$$e^x(x^2-2x)=e^2\cdot 0=0.$$

So

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

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7. Compare with options

$$y(2)=0$$

So the correct option is:

C: $0$

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8. Comparison with stored correct answer

Stored correct answer: C

Our derived answer: C

They match.