Why is the solution to a non-homogenous linear ODE written in terms of a general fundamental solution and not a matrix exponential? Generally, I see the solution to a non-homogenous linear ODE

$$ \dot{x} = Ax + b(t)\\ x(0) = x_0 $$

written as

$$ x(t) = \Phi(t)\Phi^{-1}(0)x_0 + \int_0^t \Phi(t)\Phi^{-1}(\tau)b(\tau)d\tau $$


$$ \Phi^\prime(t)=A\Phi(t). $$

At the same time, I thought $\Phi(t)=\mathbb{e}^{At}$, which means that $\Phi(0)=I$ and the above could at least be simplified to

$$ x(t) = \Phi(t)x_0 + \int_0^t \Phi(t)\Phi^{-1}(\tau)b(\tau)d\tau $$

if not further to

$$ x(t) = \mathbb{e}^{At}x_0 + \mathbb{e}^{At}\int_0^t (\mathbb{e}^{A\tau})^{-1}b(\tau)d\tau. $$

Why is it, then, that the non-homgenous solution is written as originally stated in terms of $\Phi$? Is there a change of basis or generalization that I'm not aware of, or is there a mistake in this simplication?

  • $\begingroup$ It would be problematic in non-autonomous systems. In fact, I don't know if it is possible to write $\phi(t)$ in this case $\endgroup$ – Basco Jul 23 '20 at 18:30

There's nothing wrong with using the matrix exponential in this case, where $A$ does not depend on time. However, for non-autonomous equations $\dot{x} = A(t) x + b(t)$, you can't write $\Phi(t)$ as a matrix exponential (it's sometimes called a "time-ordered exponential").


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