Thread: Sytem of diff eq View Single Post
 P: 20 This is a given system: $$D\vec{y} = A\vec{y} + \vec{b}$$ With $$A=\left\begin{array}{ccc}1&1&1\\0&2&1\\0&0&3\end{array}\right$$ And $$\vec{b}=\left\begin{array}{c}e^4^t\\0\\0\end{array}\right$$ We find $$\vec{y}_H = Y(t) \cdot \vec{c}$$ With $$Y(t)=\left\begin{array}{ccc}e^t&e^2^t&e^3^t\\0&e^2^t&e^3^t\\0&0&e^3^t\e nd{array}\right$$ Since $$\vec{y} = \vec{y}_H + \vec{y}_P$$ we still need to find is $$\vec{y}_P = Y(t) \cdot \vec{c}(t)$$ $$Y(t)$$ is already known, so whe have to find $$\vec{c}(t)$$ We know that $$D\vec{c}(t)=Y^-^1(t) \cdot \vec{b}$$ Now we are gonna replace $$Y^-^1(t)$$ by $$e^A^(^-^t^)$$ With $$e^A^(^-^t^)$$ being $$e^A^t= Y(t) \cdot Y^-^1(0)$$ So we get $$D\vec{c}(t)= e^A^(^-^t^) \cdot \vec{b}$$ My question actually is why we can replace $$Y^-^1(t)$$ by $$e^A^(^-^t^)$$ since $$e^A^t^= Y(t) \cdot Y^-^1(0)$$? P.S.: I'll post the whole excersise later if it's necessary, but I had to much trouble with the Latex code for now ;)