Initial Value Problem for a System of Linear Differential Equations

[mathmari]

We have

$$A=\begin{pmatrix}-(m_0+m_1) & m_1 & 0 \\ m_1 & -(m_1+m_2+m_3) & m_3 \\ 0 & m_3 & -(m_3+m_4)\end{pmatrix}$$

$$C=\begin{pmatrix}45m_0\\20\\35m_4\end{pmatrix}$$

So, we have to solve $X=-A^{-1}C$ subject to $z>x$, right?? (Wondering)

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I believe you're supposed to find the solution for $X = X(t)$, which contains factors $e^{\lambda t}$.
Those terms with $e^{\lambda t}$ will not and cannot be zero. (Wasntme)

We will get three relations of the form $e^{\lambda t}=0$, right?? What does this mean?? (Wondering)

 

[I like Serena]

We have

$$A=\begin{pmatrix}-(m_0+m_1) & m_1 & 0 \\ m_1 & -(m_1+m_2+m_3) & m_3 \\ 0 & m_3 & -(m_3+m_4)\end{pmatrix}$$

$$C=\begin{pmatrix}45m_0\\20\\35m_4\end{pmatrix}$$

So, we have to solve $X=-A^{-1}C$ subject to $z>x$, right?? (Wondering)

Yep. (Smile)

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We will get three relations of the form $e^{\lambda t}=0$, right?? What does this mean?? (Wondering)

I'm not so sure that's what we will get. How did you get it? (Wondering)
If we would, there would be no solution. (Wasntme)

 

[mathmari]

Yep. (Smile)

How can we find the inverse of $A$?? I got stuck right now...

 

[mathmari]

I'm not so sure that's what we will get. How did you get it? (Wondering)
If we would, there would be no solution. (Wasntme)

$\lambda$ is negative... So, that means that $t \rightarrow +\infty$, or not?? (Wondering)

 

[I like Serena]

$\lambda$ is negative... So, that means that $t \rightarrow +\infty$, or not?? (Wondering)

If $e^{\lambda t} = 0$. What is the reason you think it is? (Wondering)

 

[I like Serena]

How can we find the inverse of $A$?? I got stuck right now...

I'm not quite clear on the problem.
Aren't all $m_i$ values given? (Wondering)
Wouldn't it only be $m_1$ that we make variable?