Linear differential equations can be solved using the matrix exponential. However for upper diagonal matrices we can observe the following simplifications (given diagonal matrices commute if the $d$'s are such that if $u_{ij}$ is non-zero $d_{ii}$ and $d_{jj}$ must have the same value.):
$$ x(t)=\exp(\begin{bmatrix} d_1 & u_1 & u_2 \\ 0 & d_2 & u_3 \\ 0 & 0 & d_3 \end{bmatrix}(t-t_0))x_0 = \big(\begin{pmatrix} \exp (d_1(t-t_0)) \\ \exp (d_2(t-t_0)) \\ \exp (d_3(t-t_0)) \end{pmatrix} \cdot \exp(\begin{bmatrix} 0 & u_1 & u_2 \\ 0 & 0 & u_3 \\ 0 & 0 & 0 \end{bmatrix}(t-t_0))\big)x_0 $$ This can be simplified further since the strictly upper diagonal matrix is nil potent: $$x(t)=\Big(\begin{pmatrix} \exp (d_1(t-t_0)) \\ \exp (d_2(t-t_0)) \\ \exp (d_3(t-t_0)) \end{pmatrix} \cdot \big(\mathbb{I}_3+\begin{bmatrix} 0 & u_1 & u_2 \\ 0 & 0 & u_3 \\ 0 & 0 & 0 \end{bmatrix}(t-t_0)+\begin{bmatrix} 0 & 0 & (u_2)^2+u_1u_3 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}(t-t_0)^2\big)\Big)x_0 $$ $$x(t)=\Big(\begin{pmatrix} \exp (d_1(t-t_0)) \\ \exp (d_2(t-t_0)) \\ \exp (d_3(t-t_0)) \end{pmatrix} \cdot \begin{bmatrix} 1 & u_1(t-t_0) & u_2(t-t_0) + ((u_2)^2+u_1u_3)(t-t_0)^2 \\ 0 & 1 & u_3(t-t_0) \\ 0 & 0 & 1 \end{bmatrix}\Big)x_0 $$ This is extremely simple to solve, in particular if it is know that the degree of nilpotency is much smaller than the size of the matrix. Systems with the same degree of nil potency as this one occur in rigid body dynamics.
Does there exists an transform that transforms an arbitrary real linear ode into (one or more) upper diagonal linear (complex) ODEs whose solutions i can transform back into a solution to the arbitrary linear ODE?
The obvious idea of splitting up the matrix in an strictly upper, strictly lower and an diagonal part doesn't work as the strictly lower part doesn't commute with (diagonal+strictly upper). I'm not very familiar with matrix transforms or transforms of differential equations.