Answer: The general solution is \(y=e^{-x}(A\cos\sqrt{3}x+B\sin\sqrt{3}x)+x^2-x+2\).

First solve the associated homogeneous equation

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+2\frac{\mathrm{d}y}{\mathrm{d}x}+4y=0\).

Try \(y=e^{mx}\). Then

\(m^2+2m+4=0\).

So

\(m=\frac{-2\pm\sqrt{4-16}}{2}=-1\pm\sqrt{3}\,i\).

Hence the complementary function is

\(y_c=e^{-x}(A\cos\sqrt{3}x+B\sin\sqrt{3}x)\),

where \(A\) and \(B\) are arbitrary constants.

Now find a particular integral. Since the right-hand side is a quadratic, try

\(y_p=px^2+qx+r\).

Then

\(y_p'=2px+q\), \(y_p''=2p\).

Substitute into the differential equation:

\(2p+2(2px+q)+4(px^2+qx+r)=4x^2+8\).

Expanding gives

\(4px^2+(4p+4q)x+(2p+2q+4r)=4x^2+8\).

Equating coefficients of like powers of \(x\):

\(4p=4\Rightarrow p=1\),

\(4p+4q=0\Rightarrow 1+q=0\Rightarrow q=-1\),

\(2p+2q+4r=8\Rightarrow 2-2+4r=8\Rightarrow r=2\).

So a particular integral is

\(y_p=x^2-x+2\).

Therefore the general solution is

\(y=e^{-x}(A\cos\sqrt{3}x+B\sin\sqrt{3}x)+x^2-x+2\).