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

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

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

where

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

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?

• 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 – Basco Jul 23 '20 at 18:30

## 1 Answer

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").