This time the auxiliary equation is:

A.E. m^2-2m+4=0

Solving for m, we find that the solutions are a complex conjugate pair:

m_1=1-jsqrt3 and m_2=1+jsqrt3

The solution for our DE, using the 3rd type from the table above:

y = e^(alphax)(A\ cos\ \omega x + B\ sin\ \omega x)

we get:

y(x)= e^x(A\ cossqrt3 x+B\ sinsqrt3 x)