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If response is defined as $Res=\sum_{i=1}^{4} r_i x_i$ and $x=(x_1;x_2;x_3;x_4)$. $y$ is constant.

And

$\frac{d}{dt} x = P(y) \cdot x + Q(y)$

Then, how can we pick $r_i$'s such that the steady state Response is independent of $y$?

I have no idea how to even begin on this question. Note: $P$ and $Q$ are 4x4 and 4x1 matrices which have some elements as a function of $y$.

Additional note: $\sum x_i=1$ but we are told not to pick an obvious choice for r's. So I am pretty sure that the obvious choice is $r_i=1$ for all $i$.

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    I managed to figure it out! The thing I did was to split $\sum r_i x_i$ into the numerator and denominator, and then compare all the terms. So if say that I figured out that the sum was $\frac{r_1+r_1 r_2 y+r_3 y^2 +r_4 y^3}{1+y+y^2+y^3}$ then I can say that $r_1=k/1,r_2 r_1 = k/1, r_3=k/1, r_4=k/1$ where $k$ is any constant.2011-03-26

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Not a solution! I think this leads nowhere:
We want to eliminate $y$ from $R = \vec{r}\cdot\vec{x}$ given that $x$ follows the following differential equation:
$\dot{\vec{x}} = P(y)\cdot\vec{x} + \vec{q}(y)$
To do this, let's first eliminate the derivative with respect to time by integration:
$\vec{x}(t) = P(y)\cdot\int_0^t\vec{x}(\tau)\;d\tau + \vec{q}(y)\;t$ So we need to choose $\vec{r}$ such that the following does not depend on $y$:
$R = \vec{r}\cdot\vec{x} = \vec{r}P(y)\cdot\int_0^t\vec{x}(\tau)\;d\tau + \vec{r}\cdot\vec{q}(y)\;t$ To ensure that the above does not depend on $y$, we differentiate it with respect to $y$ and set it equal to zero. We get: $0 = \vec{r}\cdot\frac{dP(y)}{dy}\cdot\int_0^t\vec{x}(\tau)\;d\tau + \vec{r}\cdot P(y)\cdot\int_0^t\frac{\partial \vec{x}(\tau)}{\partial y}\;d\tau + \vec{r}\cdot\frac{d\;\vec{q}(y)}{dy}\;t$

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    There may be a family of solutions, but I need to find just one. ($r_i$ being constant is not an acceptable one)2011-03-26