Answer: \(y=2\sqrt{2}\,\mathrm{e}^{-x}\sin(\sqrt{2}x)+9x^2-12x+2\).

We solve

\(\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2}+2\frac{\mathrm{d}y}{\mathrm{d}x}+3y=27x^2\).

First solve the associated homogeneous equation

\(y''+2y'+3y=0\).

The auxiliary equation is

\(m^2+2m+3=0\).

So

\(m=\frac{-2\pm\sqrt{4-12}}{2}=-1\pm i\sqrt{2}\).

Hence the complementary function is

\(y_c=\mathrm{e}^{-x}\big(A\cos(\sqrt{2}x)+B\sin(\sqrt{2}x)\big)\).

For a particular integral, try

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

Then

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

\(y_p''=2p\).

Substituting gives

\(2p+2(2px+q)+3(px^2+qx+r)=27x^2\).

So

\(3px^2+(4p+3q)x+(2p+2q+3r)=27x^2\).

Equating coefficients:

\(3p=27\), \(4p+3q=0\), \(2p+2q+3r=0\).

Hence

\(p=9\), \(q=-12\), \(r=2\).

So the general solution is

\(y=\mathrm{e}^{-x}\big(A\cos(\sqrt{2}x)+B\sin(\sqrt{2}x)\big)+9x^2-12x+2\).

Now use the conditions.

When \(x=0\), \(y=2\):

\(2=A+2\), so \(A=0\).

Differentiate:

\(y'=\mathrm{e}^{-x}\big(-\sqrt{2}A\sin(\sqrt{2}x)+\sqrt{2}B\cos(\sqrt{2}x)\big)-\mathrm{e}^{-x}\big(A\cos(\sqrt{2}x)+B\sin(\sqrt{2}x)\big)+18x-12\).

When \(x=0\), \(y'=-8\):

\(-8=\sqrt{2}B-A-12\).

Since \(A=0\),

\(\sqrt{2}B=4\Rightarrow B=2\sqrt{2}\).

Therefore the required particular solution is

\(y=2\sqrt{2}\,\mathrm{e}^{-x}\sin(\sqrt{2}x)+9x^2-12x+2\).