Answer: \(y=\mathrm{e}^x(9x-4)-2\sin x\).

We solve

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

with \(y=-4\) and \(\dfrac{\mathrm{d}y}{\mathrm{d}x}=3\) when \(x=0\).

First solve the homogeneous equation

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

Its auxiliary equation is

\(m^2-2m+1=0\),

so \((m-1)^2=0\).

This gives a repeated root \(m=1\), so the complementary function is

\(y_c=\mathrm{e}^x(Ax+B)\).

For a particular solution, try

\(y_p=p\sin x+q\cos x\).

Then

\(y_p'=p\cos x-q\sin x\),

\(y_p''=-p\sin x-q\cos x\).

Substitute into the differential equation:

\((-p\sin x-q\cos x)-2(p\cos x-q\sin x)+(p\sin x+q\cos x)=4\cos x\).

Collecting coefficients gives

\(2q\sin x-2p\cos x=4\cos x\).

So

\(2q=0\Rightarrow q=0\),

\(-2p=4\Rightarrow p=-2\).

Hence

\(y_p=-2\sin x\).

So the general solution is

\(y=\mathrm{e}^x(Ax+B)-2\sin x\).

Differentiate:

\(y'=A\mathrm{e}^x+\mathrm{e}^x(Ax+B)-2\cos x\).

Now use the initial conditions.

At \(x=0\),

\(y(0)=B=-4\), so \(B=-4\).

Also

\(y'(0)=A+B-2=3\).

Substituting \(B=-4\):

\(A-6=3\Rightarrow A=9\).

Therefore the required particular solution is

\(\boxed{y=\mathrm{e}^x(9x-4)-2\sin x}.\)