Given $$x' = \pmatrix{ -3t^2 & 0 \\ 1 & -3t^2 }x $$
What is the fundamental matrix of this system?
I decomposed the matrix into a nilpotent $\pmatrix{ 0 & 0 \\ 1 & 0 }$ and $e^{-3t^2} I$, i.e, which gave me $$e^{At}x_0 = e^{-3t^3}\pmatrix{ 1 & 0 \\ t & 1 }x_0$$ for the fundmental matrix.
Where have I gone wrong? Am I not using the definition properly?
Thanks in advance.
You've made a slight mistake. Consider that for $x'(t) = A(t)x(t)$, the solution is $x(t) = e^{\int_0^t A(t')\,\mathrm{d}t'}x(0)$. We have $$\int_0^t \begin{pmatrix} -3t'^2 & 0 \\ 1 & -3t'^2 \end{pmatrix}\mathrm{d}t' = \begin{pmatrix} -t^3 & 0 \\ t & -t^3 \end{pmatrix}$$ which has the diagonal component $$D = \begin{pmatrix} -t^3 & 0 \\ 0 & -t^3 \end{pmatrix} = -t^3I$$ and the nilpotent component $$N = \begin{pmatrix} 0 & 0 \\ t & 0 \end{pmatrix}$$ As $D$ and $N$ commute, we have that $$e^{D+N} = e^De^N = (e^{-t^3}I)\begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix} = e^{-t^3}\begin{pmatrix} 1 & 0 \\ t & 1 \end{pmatrix}$$