Answer: The particular solution satisfying the given conditions is

\(y=\mathrm{e}^{-x/2}\left(-13\cos\frac{x}{2}+13\sin\frac{x}{2}\right)+4x^2-13x+13\).

We solve

\(2\frac{\mathrm{d}^2y}{\mathrm{d}x^2}+2\frac{\mathrm{d}y}{\mathrm{d}x}+y=4x^2+3x+3\),

with \(y=0\) and \(\frac{\mathrm{d}y}{\mathrm{d}x}=0\) when \(x=0\).

First solve the homogeneous equation

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

Its auxiliary equation is

\(2m^2+2m+1=0\).

So

\(m=\frac{-2\pm\sqrt{4-8}}{4}=-\frac12\pm\frac12 i\).

Hence the complementary function is

\(y_c=\mathrm{e}^{-x/2}\left(A\cos\frac{x}{2}+B\sin\frac{x}{2}\right)\).

For a particular integral, try

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

Then

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

\(y_p''=2p\).

Substituting into the differential equation gives

\(2(2p)+2(2px+q)+(px^2+qx+r)=4x^2+3x+3\),

so

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

Equating coefficients:

\(p=4\),

\(4p+q=3\),

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

Thus

\(q=-13\), \(r=13\).

So the general solution is

\(y=\mathrm{e}^{-x/2}\left(A\cos\frac{x}{2}+B\sin\frac{x}{2}\right)+4x^2-13x+13\).

Now use the conditions.

From \(y(0)=0\):

\(A+13=0\), so \(A=-13\).

Differentiate:

\(y'=\mathrm{e}^{-x/2}\left(-\frac12A\sin\frac{x}{2}+\frac12B\cos\frac{x}{2}\right)-\frac12\mathrm{e}^{-x/2}\left(A\cos\frac{x}{2}+B\sin\frac{x}{2}\right)+8x-13\).

Using \(y'(0)=0\):

\(\frac12B-\frac12A-13=0\).

Substitute \(A=-13\):

\(\frac12B+\frac{13}{2}-13=0\), so \(B=13\).

Therefore

\(y=\mathrm{e}^{-x/2}\left(-13\cos\frac{x}{2}+13\sin\frac{x}{2}\right)+4x^2-13x+13\).