View Single Post
Jan10-04, 03:27 PM
P: 20
This is a given system: [tex]D\vec{y} = A\vec{y} + \vec{b}[/tex]
With [tex]A=\left\begin{array}{ccc}1&1&1\\0&2&1\\0&0&3\end{array}\right[/tex]
And [tex]\vec{b}=\left\begin{array}{c}e^4^t\\0\\0\end{array}\right[/tex]

We find [tex]\vec{y}_H = Y(t) \cdot \vec{c}[/tex]

With [tex]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[/tex]

Since [tex]\vec{y} = \vec{y}_H + \vec{y}_P[/tex] we still need to find is [tex]\vec{y}_P = Y(t) \cdot \vec{c}(t)[/tex]
[tex]Y(t)[/tex] is already known, so whe have to find [tex]\vec{c}(t)[/tex]
We know that [tex]D\vec{c}(t)=Y^-^1(t) \cdot \vec{b}[/tex]

Now we are gonna replace [tex]Y^-^1(t)[/tex] by [tex]e^A^(^-^t^)[/tex]

With [tex]e^A^(^-^t^)[/tex] being [tex]e^A^t= Y(t) \cdot Y^-^1(0)[/tex]
So we get [tex]D\vec{c}(t)= e^A^(^-^t^) \cdot \vec{b}[/tex]

My question actually is why we can replace [tex]Y^-^1(t)[/tex] by [tex]e^A^(^-^t^)[/tex] since [tex]e^A^t^= Y(t) \cdot Y^-^1(0)[/tex]?

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 ;)
Phys.Org News Partner Science news on
World's largest solar boat on Greek prehistoric mission
Google searches hold key to future market crashes
Mineral magic? Common mineral capable of making and breaking bonds