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Let $A$ be a constant $n\times n$ matrix . Consider the system the system $\left\{\begin{array}{cc}x'&= Ax\\x(0)&=x_0 \end{array}\right..$

$\quad\rm(a)$ Prove that the solution can be expressed in the form $\mathcal{L}_\lambda^{-1}\left\{{\left( {\lambda I-A} \right)^{-1}}\right\}(t)\;x_0$.

Here I can only differentiate and check, but I don't know how to differentiate this. )= Please show me how. And I don't know - what is that lambda? :S

$\quad\rm(b)$ Show that $$\mathcal{L}\left\{{e^{At}}\right\}(\lambda)=\left({\lambda I-A} \right)^{-1}.$$

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What is $L$? The system $x'=Ax, x(0)=x_0$ has the unique solution $x(t)=e^{At}x_0$. –  Jeff Nov 8 '11 at 21:05
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Note: $\mathcal{L}\{f\;\}(\lambda)$ is the Laplace transform of $f(t)$ and $(\lambda I-A)^{-1}$ is a matrix-valued function of $\lambda$.


Apply the Laplace transform $\mathcal{L}$ to both sides of the system given by $x'=Ax$ using the rule for derivatives (among other rules) and we obtain

$$\lambda\,\mathcal{L}\{x\}(\lambda)-x(0)=A\mathcal{L}\{x\}(\lambda)$$

$$(\lambda I-A)\mathcal{L}\{x\}(\lambda)=x_0$$

$$x(t)=\mathcal{L}_\lambda^{-1}\left\{(\lambda I-A)^{-1}\right\}(t)\;x_0. $$

And for part $\rm(b)$ we have

$$\mathcal{L}\left\{\exp(At)\right\}(\lambda)=\int_0^\infty e^{At}e^{-\lambda t}dt $$

$$=\int_0^\infty e^{-(\lambda I-A)t}dt=\left[-(\lambda I-A)^{-1}e^{-(\lambda I-A)t}\right]_0^\infty =(\lambda I-A)^{-1}.$$

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