I need some feedback to complete my solution for the following exercise:
Given a matrix $ A\in\mathbb{R}^{d\times d}$ and a vector $ b\in\mathbb{R}^{d}$, take a look at the initial value problem (IVP)
$$ y'=Ay+b\quad\land\quad y(0)=y_{0}\in\mathbb{R}^{d}$$
Determine the solution for the IVP by using variation of parameters. Calculate the occurring integral explicitly in case of $ b\in\operatorname{img}(A)$.
Well, one can find $y(x)=\mathrm{e}^{xA}y_{0}$ as solution for the homogeneous differential equation. Now using the method of variation of parameters by inserting $ y(x)=\mathrm{e}^{xA}y_{0}(x)$ in the system and knowing that the matrix exponential is invertible
\begin{align} &&A\mathrm{e}^{xA}y_{0}(x)+\mathrm{e}^{xA}y_{0}'(x) &= A\mathrm{e}^{xA}y_{0}(x)+b \\ \Leftrightarrow && \mathrm{e}^{xA}y_{0}'(x) &= b\\ \Leftrightarrow && y_{0}'(x) &=\mathrm{e}^{-xA}b \\ \Leftrightarrow && y_{0}(x) &= y_{0} + \left(\int_{0}^{x}\mathrm{e}^{-tA}b\mathrm{d}t\right) \end{align} How to go on from that point onwards? Can I do something like that or? \begin{align} \overset{?}{\Rightarrow} && y_{0}(x) &= y_{0} + \left(\int_{0}^{x}\mathrm{e}^{-tA}\mathrm{d}t\right)b \\ \Leftrightarrow && y_{0}(x) &= y_{0}+A^{-1}b\mathrm{e}^{-xA} \end{align}
Thanks in advance.
You stumbled evaluating the integral in the last step. $$\eqalign{ d\left(e^{-tA}\right) &= -Ae^{-tA}dt \\ e^{-xA}-I &= \int_{0}^{x}-Ae^{-tA}dt \\ A^{-1}(I-e^{-xA}) &= \int_{0}^{x}e^{-tA}dt \\ }$$ Using the function $$\phi(x) = \frac{e^x-1}{x}$$ this result can be written as $$\eqalign{ \int_{0}^{x}e^{-tA}dt &= \phi(-xA)\;x \\ }$$ And your variational approach yields $$\eqalign{ y_0(x) &= y_0 + \phi(-xA)\left(xb\right) \\ y(x) &= e^{xA}y_0 + \phi(xA)\left(xb\right) \\ }$$