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Reason
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#1
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 ;)
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