Solution to general nonhomogenous time invariant system of ODEs

Question

I am just looking for an explanation to why the solution to this matrix equation

$$\frac{du(t)}{dt}=Au(t)+b,\quad u(0) = u_0,\quad u,b\in R^n\quad A\in R^{n\times n}$$

is given by, if $A$ is invertible,

$$u(t) = \exp(At)u_0+A^{-1}(\exp(At)b-b).$$

I know this is the correct solution, inserting into the ODE system confirms that. But I would just like a more satisfactory derivation of it. I was thinking that the general solution to this problem was given by

$$u(t) = \Phi (t,0)u_0+\int^{t}_{0}\Phi (t,s)bds$$

where I have used some of my knowledge from systems theory. Here $\Phi(t,t_0)$ is the transition matrix given in this case by

$$\Phi(t,t_0=0)=\exp(A(t-t_0))=\exp(At).$$

So the solution is given by (according to me!)

$$u(t) = \exp(At)u_0+\int^{t}_{0}\exp(A(t-s))bds.$$

Where do I go wrong? Or these equivalent presentations?

Best regards//

Answer 1

You did nothing wrong. Note that $$\int^{t}_{0}e^{A(t-s)}b\,ds=-A^{-1}e^{A(t-s)}b\bigg|^{s=t}_{s=0}=-A^{-1}b+A^{-1}e^{At}b.$$ So indeed $$u(t) = e^{At}u_0+\int^{t}_{0}e^{A(t-s)}b\,ds=e^{At}u(0)+A^{-1}(e^{At}b-b).$$

Answer 2

Another way,

As the DE is linear we have $u = u_h + u_p$ with

$$ \cases{ \dot u_h = A u_h\\ \dot u_p = A u_p + b } $$

now

$$ u_h = e^{A t}c $$

and for the particular we propose $u_p = e^{A t}c(t)$. Substituting now into the particular we have

$$ A e^{A t}c(t) + e^{A t}\dot c(t) = A e^{A t}c(t) + b\Rightarrow e^{A t}\dot c(t) = b $$

and now if $A$ is invertible

$$ c(t) = -A^{-1}e^{-A t}b\Rightarrow u_p = -e^{A t}A^{-1}e^{-A t}b $$

and finally

$$ u = e^{A t}u_0 - e^{A t}A^{-1}e^{-A t}b\Leftrightarrow e^{At}u_0+\int^{t}_{0}e^{A(t-s)}bds. $$