\(\frac{dy}{dx}+y \cos x= 4\cos x\)

That is \(\frac{dy}{dx}+(\cos x)y= 4\cos x\)

\(\frac{dy}{dx}+P(x)=Q(x)\)

So \(P(x)= \cos x\), \(Q(x)= 4\cos x\)

Integrating factor is

\(I.F.= e^{(\int P(x)dx)}\)

\(= e^{\int \cos xdx}\)

\(= e^{\sin x}\)

\(y \cdot I.F.= \int Q(x) \cdot I.F. dx+c\)

\(ye^{\sin x}= \int 4\cos x e^{\sin x} dx+c\)

\(= 4 \int e^{\sin x} \cos x dx+c\)

\(= 4 \int e^t dt+c\) (by subtitution)

\(= 4e^t+c\)

\(ye^{\sin x}= 4e^{\sin x}+c\)

\(y(x)=4e^{\sin x}+c\)

Now apply the initial condition y(0)=6 in the general solution

\(4e^{\sin(0)}+c=6\)

\(4e^0+c=6 => 4+c=6\)

c=6-4=2 Now substitute 2 for c in general solution

Thus,the particular solution is \(y(x)=4e^{\sin x}+2\)