First, the characteristic solution to the homogeneous part:
[tex]\dfrac{\mathrm d^2y}{\mathrm dx^2}-2\dfrac{\mathrm dy}{\mathrm dx}+4y=0[/tex]
has characteristic equation
[tex]r^2-2r+4=0[/tex]
with roots [tex]r=1\pm\sqrt3i[/tex]. So the characteristic solution has the form
[tex]y_c=e^x(C_1\cos\sqrt3x+C_2\sin\sqrt3x)[/tex]
Now for the nonhomogeneous part. Using the method of undetermined coefficients, we can try a particular solution of the form
[tex]y_p=e^x(a\cos x+b\sin x)[/tex]
which has derivatives
[tex]\dfrac{\mathrm dy_p}{\mathrm dx}=e^x((a+b)\cos x+(-a+b)\sin x)[/tex]
[tex]\dfrac{\mathrm d^2y_p}{\mathrm dx^2}=2e^x(b\cos x-a\sin x)[/tex]
Substituting [tex]y_p[/tex] and its derivatives into the equation gives
[tex]2e^x(b\cos x-a\sin x)-2e^x((a+b)\cos x+(-a+b)\sin x)+4e^x(a\cos x+b\sin x)=e^x\cos x[/tex]
[tex]2e^x(a\cos x+b\sin x)=e^x\cos x[/tex]
which means
[tex]\begin{cases}2a=1\\2b=0\end{cases}\implies a=\dfrac12,b=0[/tex]
so that the particular solution must be
[tex]y_p=\dfrac12e^x\cos x[/tex]
Now the general solution will be
[tex]y=y_c+y_p[/tex]
[tex]y=e^x(C_1\cos\sqrt3x+C_2\sin\sqrt3x)+\dfrac12e^x\cos x[/tex]
