"Simple" Linear ODE of order 100 I'm trying to solve the following ODE
            $$y^{(100)} + 100y = 0,$$
and I get the characteristic polynomial $\lambda^{100} + 100$, but do not know how to find its roots. Is there a way to find a general formula for all the roots or do I have to solve the equation using another method?
The answer is 
$$y(t) = \sum_{n=0}^{99} c_n e^{r_n t},$$ where $r_n = (100)^\frac{1}{100}e^{\frac{2n+1}{100}\pi i} $.

Look for solutions of the form $\mathrm{e}^{s t}$. The ODE then becomes
$$(s^{100} + 100) \, \mathrm{e}^{s t} = 0$$
Hence, $s^{100} + 100 = 0$, or,
$$s^{100} = 100 \, \mathrm{e}^{i (2k+1) \pi}$$
where $k \in \mathbb Z$. Taking the 100th root, we obtain $100$ solutions
$$\{ \sqrt[100]{100} \, \mathrm{e}^{i (2k+1) \frac{\pi}{100}} \mid k \in \{0,1,\dots,99\} \}$$
