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?