Answer: \(k=2\).

The general solution is \(y=(A+Bx+2x^2)e^{2x}\).

The particular solution is \(y=(3-8x+2x^2)e^{2x}\).

Let us first find a suitable particular integral of the form \(y=kx^2e^{2x}\).

Differentiating,

\(y'=k(2xe^{2x}+2x^2e^{2x})\)

and

\(y''=k(2e^{2x}+8xe^{2x}+4x^2e^{2x})\).

Substitute into \(y''-4y'+4y=4e^{2x}\):

\(k(2e^{2x}+8xe^{2x}+4x^2e^{2x})-4k(2xe^{2x}+2x^2e^{2x})+4kx^2e^{2x}=4e^{2x}\).

The \(xe^{2x}\) and \(x^2e^{2x}\) terms cancel, leaving

\(2ke^{2x}=4e^{2x}\), so \(k=2\).

Hence a particular integral is \(2x^2e^{2x}\).

For the homogeneous equation \(y''-4y'+4y=0\), the auxiliary equation is

\(m^2-4m+4=0\), so \((m-2)^2=0\).

Thus the complementary function is

\(y=(A+Bx)e^{2x}\).

So the general solution is

\(y=(A+Bx+2x^2)e^{2x}\).

Now use the conditions \(y=3\) and \(\frac{dy}{dx}=-2\) when \(x=0\).

At \(x=0\),

\(y(0)=A=3\).

Differentiate \(y=(A+Bx+2x^2)e^{2x}\):

\(y'=(B+4x)e^{2x}+2(A+Bx+2x^2)e^{2x}\).

So at \(x=0\),

\(y'(0)=B+2A\).

Given \(y'(0)=-2\) and \(A=3\),

\(B+6=-2\), hence \(B=-8\).

Therefore the particular solution is

\(y=(3-8x+2x^2)e^{2x}\).