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)`