Linear homogeneous ODE with time dependent matrix I was wondering, if $x'(t)=A(t)x(t)$ where $A(t)$ is a continuos $n\times n$ matrix, how do I show that it has $n$ linearly independent solutions?
Think of $x(t)$ and $A(t)$ as 1-dimensional real functions, then it's a standtard ODE whose solution is $$x(t)=c e^{\int A(t)dt}$$

where $c\in \mathbb R$. This is actually the solution for your case as well, but now $c \in \mathbb R^n$ and all that remains is to know how to compute $e^{\int A(t)dt}$. I bet you already guess how to compute $\int A(t)dt$, all that remains then is to find out how to compute $e^M$ being $M$ a matrix. That is called the matrix exponential, and you can find how to compute it in Wikipedia. In fact, you can compute $f(M)$ for any function $f$

Also, since $A(t)$ is continous, the matrix $\int A(t)dt$ will always exist, and the following relation for the determinant of the matrix $e^{\int A(t)dt}$ will hold $det(e^{\int A(t)dt})=e^{tr(\int A(t)dt)}$ by Jacobi's formula. So you can already compute how the determinant of $e^{\int A(t)dt}$ will vary with time without actually computing the matrix, and as you can see, it will never be $0$, which tells you that the equation will have $n$ independent solutions.
Think of $x(t)$ and $A(t)$ as 1-dimensional real functions, then it's a standtard ODE whose solution is $$x(t)=c e^{\int A(t)dt}$$

where $c\in \mathbb R$. This is actually the solution for your case as well, but now $c \in \mathbb R^n$ and all that remains is to know how to compute $e^{\int A(t)dt}$. I bet you already guess how to compute $\int A(t)dt$, all that remains then is to find out how to compute $e^M$ being $M$ a matrix. That is called the matrix exponential, and you can find how to compute it in Wikipedia. In fact, you can compute $f(M)$ for any function $f$

Also, since $A(t)$ is continous, the matrix $\int A(t)dt$ will always exist, and the following relation for the determinant of the matrix $e^{\int A(t)dt}$ will hold $det(e^{\int A(t)dt})=e^{tr(\int A(t)dt)}$ by Jacobi's formula. So you can already compute how the determinant of $e^{\int A(t)dt}$ will vary with time without actually computing the matrix, and as you can see, it will never be $0$, which tells you that the equation will have $n$ independent solutions.