Variable coefficient systems & Matrix exponential in differential equation?

Question

So, I'm pretty familiar with solving differential equations $\frac{dx}{dt}=Bx$, in which $B$ is an $n \times n$ matrix with an initial value of $x(0)=x_0$. I believe I would get a solution along the lines of $x(t) = e^{tB}x_0$. However, now I have the following problem:

$\frac{dx}{dt}=e^{tB}x$ with initial condition $x(0)=x_0$.

Using the suggested protocol, I did the following:

I substituted $e^{tB}$ with $f(t)$.

$\frac{dx}{dt}=f(t)x$

Using the separable equations method:

$\frac{dx}{x}=f(t)dt$

$\int\frac{dx}{x}=\int f(t)dt$

$ln|x| + C = \int f(t)dt$

As suggested below, raising both sides to the power of $e$ and using A as a constant:

$x = Ae^{\int f(t)dt}$.

Now I substituted back in $e^{tB}$ in the solution and solve the integral using the initial values provided:

$x = Ae^{\int_{0}^{t} e^{\tau B}d\tau}$.

By integrating, I got the solution $x(t) = Ae^{(\frac{1}{B}e^{tB}-\frac{1}{B})}$

Answer 1

That is not a second order equation. It is just $x' = f(t)x$ for a different $f(t)$ than in your first equation.

If we were dealing with functions on the real numbers, this would be easy to solve: $$\frac {dx}x = f(t) dt$$ $$\ln x + C = \int f(t) dt$$ $$ x = Ae^{\int f(t)dt}$$ for some constant $A$.

Do you need any more guidance than that for the matrix case?

Answer 2

In general, the formula $x(t)=e^{\int_{0}^tf(u)du}x_0$ does not work.

Indeed, we would like to write $x'(t)=f(t) e^{\int_{0}^t f(u)du}x_0=f(t)x(t)$. Ufortunately, it is, in general, false because $\int_{0}^t f(u)du$ and $f(t)$ do not commute.

Fortunately, here $f(t)=e^{tB}$ commutes with $\int_{0}^t f(u)du=(e^{tB}-I)B^{-1}$ (of course, we assume that $B$ is invertible).