Since you write $c_1 = \cdots = c_n = \frac 1n$, I suspect that you're referring to the solution formula that's something like
$$
y = c_0 e^x + c_1 e^{\omega x} + \cdots + c_{n-1}e^{\omega^{(n-1)} x},
$$
where $\omega = e^{2 \pi i/n}$, as opposed to the more typical expression involving sines and cosines.
It depends what you mean by "doing this". If you want to simply show that taking $c_0 = c_1 = \cdots = c_{n-1} = 1/n$ solves the problem, then that's straightforward: of course, we find that
$$
y(0) = (1/n)e^0 + \cdots + (1/n)e^{0} = n \cdot (1/n) = 1.
$$
On the other hand, for $k = 1, \dots, n-1$, the formula for the sum of a geometric series yields
\begin{align}
y^{(k)} &= (1/n) + (1/n)\omega^k e^0 + \cdots + (1/n)\omega^{k(n-1)}e^{0}
\\ & = \frac 1n \sum_{j=0}^{n-1} \omega^{jk} = \frac 1n \cdot \frac{1 - (\omega^k)^n}{1 - \omega^k} = \frac 1n \cdot \frac{0}{1 - \omega^k} = 0.
\end{align}
On the other hand, if you're looking for a systematic way to solve these initial value problems, it turns out that the solution process can be made fairly quick. Given initial values
$$
y(0) = y_0, \quad y'(0) = y_0', \quad \dots\quad ,\quad y^{(n-1)}(0) = y^{(n-1)}_0,
$$
we find that the system of equations over $c_0,\dots,c_{n-1}$ can be written as
$$
\pmatrix{
1 & 1 & \cdots & 1\\
1 & \omega & \cdots & \omega^{n-1}\\
\vdots & \vdots & \ddots & \vdots\\
1 & \omega^{n-1} & \cdots & \omega^{(n-1)^2}}\pmatrix{c_0 \\ c_1 \\ \vdots \\ c_{n-1}} = \pmatrix{y_0 \\ y_1 \\ \vdots \\ y_{n-1}}.
$$
The coefficient matrix here turns out to be $\sqrt{n} \cdot W$, where $W$ denotes the DFT matrix of size $n$. A key property of $W$ is that it is a unitary matrix, which means that its inverse is particularly easy to compute. With that established, we note that
$$
(\sqrt{n} \cdot W)^{-1} = \frac 1{\sqrt{n}} W^* = \frac 1{\sqrt{n}} \overline{W},
$$
where $W^*$ denotes the conjugate-transpose of $W$ and $\overline{W}$ simply means that we take the conjugate of all entries. Notably, we have $\overline{\omega^k} = \omega^{-k}$ for all $k$. Applying that and going through the matrix multiplication in $\mathbf c = (\sqrt{n} \cdot W)^{-1} \mathbf y$ yields the neat formula
$$
c_k = \frac 1n \sum_{j=0}^{n-1} \omega^{-jk}\, y_j.
$$
