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2. Hemodialysis is a process by which a machine is used tofilter urea and other waste products from a patient's blood ifthe kidneys fail. The amount of urea within a patient duringdialysis is sometimes modeled by supposing there are twocompartments within the patient: the blood, which isdirectly filtered by the dialysis machine, and another com-partment that cannot be directly filtered but that is con-nected to the blood. A system of two differential equationsdescribing this is$$\frac{d c}{d t}=-\frac{K}{V} c+a p-b c \quad \frac{d p}{d t}=-a p+b c$$where $c$ and $p$ are the urea concentrations in the blood andthe inaccessible pool (in $\mathrm{mg} / \mathrm{mL} )$ and all constants arepositive (see also Exercise 14 in the Review Section of thischapter). Suppose that $K / V=1, a=b=\frac{1}{2},$ and the initialurea concentration is $c(0)=c_{0}$ and $p(0)=c_{0} \mathrm{mg} / \mathrm{mL}$$$\begin{array}{l}{\text { (a) Classify the equilibrium of this system. }} \\ {\text { (b) Solve this initial-value problem. }}\end{array}$$

$\frac{1}{2} c_{0} e^{\left(-1+\frac{\sqrt{2}}{2}\right) t} \left[ \begin{array}{c}{1} \\ {1+\sqrt{2}}\end{array}\right]+\frac{1}{2} c_{0} e^{\left(-1-\frac{\sqrt{2}}{2}\right)} t \left[ \begin{array}{c}{1} \\ {1-\sqrt{2}}\end{array}\right]$

Calculus 3

Chapter 10

Systems of Linear Differential Equations

Section 3

Applications

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So we have the following system and first we want to classify the equilibrium. So in order to do so, we need to rewrite this stone in matrix form. So we have, um e c e d T and dp dt which we can just call the ex d. T as a vector. And this is equal to some matrix times. Uh, c p because those are labels. And so it's filling this matrix. Um, so let's rewrite D c e d t. Since we're given that A and B, you're 1/2 and k over V is one. Ah, this is just equal to negative C plus 1/2 p minus 1/2 see, or 1/2 p minus 3/2. See, So you can feel that in our matrix here s O. P. Needs to be multiplied by 1/2 and see he needs to be multiplied by negative. We could be right. Our second equation also as negative 1/2 p plus 1/2 seat soapy needs to be multiplied by negative 1/2 and see needs to be multiplied by positive 1/2. So the sir matrix. So we need to find I convey I use and I can factors. Um, so we've done this a lot by this point, so we just find the determinant of Ah, this matrix a minus lambda I and we get the Lambda. Wanted to are equal to negative one closer, minus square to two over too. And then we saw the system. Um, a man is ling chi times the equal zero to get our wagon vectors, which come out to be, um, one and one plus or minus square to two. So to classify the equilibrium, which is what this question is all about, um r Agon values rewrite those reminder we have that they are the same sign which tells us that we have a node and this What is their sign? Their signs Negative. So this tells us that we have a stable equilibrium So we have a stable node now for party. We need to solve the initial body problem. So we know our general solution is equal to some constant times Eat of the land that won t So we have negative one plus squared a two over to tea times. Corresponding. I am vector one and one poor square to two, plus another constant heat of the Lambda to tea. So we have minus and corresponding Aiken vector with a minus sign. Don't order us all for the C one and C two. We need to use the initial condition. So using the initial condition, um, X zero is equal. Thio, uh, CEO zero times p 00 which are both given to be, uh, these value. See? Not so if the actually plug zero into our general solution. We see that exponential go away. So we get, um, c one times the first Aiken vector plus C two times the second in vector. So this tells me that, um, c one plus c two equals this sea, not value. Those is just, uh, see nauseous. The number do you want to see? Two are actually constants that we're solving for. So that's the first equation. And second equation tells me that one full square to two. So you want close? One minus squared. A juicy too asked equal zero, do you not? So now let's, um, substitute C two equals c not minus seat one into this equation. So we're gonna leave the one postcard to to see one plus one minus square to two instead of C two. I'm gonna put see, not mine issue on. And this is equal to see Not so let's combine our coefficients off. See one. So I have one plus square to two. And then I also have, um, a minus this. So we have the opposite of that. So we have negative one plus square to two, and this is all multiplied by C one. And we have, um, C zero on the other side. So if we move the sea, not to the other side, we have the opposite of this again. So negative one plus squared too. Plus the one we already have on the right hand side. So one in minus one. Cancel one minus one. Cancel here. So we get to square roots or to see one equals square to to see. Not so solving. For C one, we could see one equals square root of two over two square roots or 20 square. It's cancelled. So we get C one equals 1/2 seen hot. So we have this now we're gonna plug it in over here to get seats, so it's see, not minus C one, which is 1/2. See? Not so we get 1/2. See, Not again. So C one and C two are both equal to each other. So now all we need to do is substitute those in for our general solution. So I'm gonna get rid of, uh, you wanted to see to hear. And we can substitute in our constants we found So see, one is 1/2 see, not and see to is the same value. So this is our final solution to the initial value problem.

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