Fourth root of unknown positive constant-4th order DE

Question

I am attempting to solve a fourth order homogeneous linear differential equation: $${d^4y\over dx^4}-ay=0$$ The auxiliary equation is $$m^4-a=>m^4=a$$ But I don't how to find the roots of $$\sqrt[4]{a}$$ where 'a' is an unknown positive constant. I would appreciate your help in this regard.

Answer 1

As Omnomnomnom has expressed the values obtained from the equation $m^{4} = a$ are $m \in \{\sqrt[4]{a}, - \sqrt[4]{a}, i \sqrt[4]{a}, -i \sqrt[4]{a} \}$. The application of these values to the differential equation is the following. \begin{align} y^{(4)} - a y = 0 \end{align} has solutions of the form $y = e^{\alpha x}$ and leads to the equation $\alpha^{4} = a$. From this the solutions is of the form \begin{align} y(x) &= A \, e^{\sqrt[4]{a} x} + B \, e^{-\sqrt[4]{a} x} + C \, e^{i\sqrt[4]{a} x} + D \, e^{-i\sqrt[4]{a} x} \end{align} or \begin{align} y(x) &= A \, \cosh(\sqrt[4]{a} x) + B \, \sinh(\sqrt[4]{a} x) + C \, \cos(\sqrt[4]{a} x) + D \, \sin(\sqrt[4]{a} x). \end{align}