Originally Posted by

**bakinbacon** the problem is:

y^{(4)}-5y''-36y=0

=> m^{4}-5y^{2}-36=0

=> (m^{2}-9)(m^{2}+4)=0

therefore, m_{1}=3, m_{2}=-3, m_{3}=2i, m_{4}=-2i.

I have learned how to find the general solution for cases with distinct real roots and for cases with complex roots but I am unsure how to I would go about finding the general solution for a linear equation that has both real and complex roots.

Would the general solution be the following

c_{1}e^{3}^{x}+c_{2}e^{-3x}+c_{3}(cosx+(2i)sinx)+c_{4}(cosx+(-2i)sinx)

by using euler's formula?