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Populations of aphids and ladybugs are modeled by…

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Problem 9 Medium Difficulty

In Example 1(b) we showed that the rabbit and wolf populations satisfy the differential equation
$ \frac {dW}{dR} = \frac {-0.02W + 0.00002RW}{0.08R - 0.001RW} $
By solving this separable differential equation, show that
$ \frac {R^{0.02}W^{0.08}}{e^{0.00002R}e^{0.001W}} = C $
where $ C $ is a constant.
It is impossible to solve this equation for $ W $ as an explicit function of $ R $ (or vice versa). If you have a computer algebra system that graphs implicitly defined curves, use this equation and your CAS to draw the solution curve that passes through the point (1000, 40) and compare with Figure 3.

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Hi, guys. Rich in chapter nine, Section six, Problem nine in which were given a differential equation that we need to solve and rearrange isolating for C and then we need to graph it through the point of 1040. So first things first solving differential equation requires three steps. When you separate, the variable is getting the W's onto one side, the our zone, toe, other side, and then integrate both sides and then finally isolate for on arbitrary constancy. So, first of all, we're going to eliminate the denominators which will help us eventually separate the variables so by so we're gonna multiply the denominators on each side. Okay, Now that we've eliminated, the denominators are variables. They're still not separated. Whoever we can see through observation that you could factor out are on this left hand side and w from the right hand side, which will allow us to then separate the variables and get them on the respective sides. So we're gonna quickly do that. Now we're gonna factor out the w over here. Okay, so now you can see that no term has both variables together. So now we just need to simply put the variables on the right side, since right integrating, we need all the W's with the DW all the R's on the side of the d r. So in order to achieve this, we're going to divide each side by r. W, allowing us to cancel them out and get them on the right side. So R. W So we're gonna cancel the arson, this side and also you can see that there's a W in the second term of the left hand side to become also so that out some believe we're gonna divide by our w here. And we're also able to kids out this are when we write the equation like so So this is over dope for you. Since we're able to cancel out, the W in this term can simply have minus 00 0.1 And then, since we divided by r were able to put this over, are over. Sorry, Since we cancelled out there on this term, we can simply have 0.2 Remember, he's in brackets now. The variables are separated and can be integrated, so we're going to simply thrown into role. So it's are integrating. So since one over W Integrated is just Ellen of W this our first term is gonna be 0.8 Ellen of W. Minus 0.1 times w plus C one, which is just some arbitrary, constant. And then similarly, on the right hand side, we're going to get negative 0.2 times Ellen off our plus 0.0 to r plus C two, which is going to some arbitrary constant. So if you go back to the equation at the top that we need A that we need to transform a nicely for see, you'll see that there's no l ends here. So in order to get rid of Dylan's, let's move them all into inside everything else on the other side. So we're gonna have 0.0 a Ln of w. And let's move the Ellen of our onto the sides plus 0.2 Ellen of our and then we're gonna have all the confidence on the other side. So not constant, but all the other terms to our that's on the right hand side now, plus 0.1 w, which were left inside. And then let's to have plus C three. So in order instead of having C two and C one with your arbitrary constants, just combine those so we'll have si three equals C to minus C. One. This was caused from moving. See one from the left hand side to the right hand side. Oh, just keep it simpler if we have one constant term. Okay, so now let's look at properties of algorithms to understand how we can simplify the left hand side. So the first property is if we have some constant a times Allen of X that we can also then rewrite that as Ellen of X to the A. So on the left hand side, both doesn't have called. Both terms have a constant, which we now can just move as an exponents inside that not for longer than secondly, if we have Ellen of X plus Ellen off Ellen off, why that will equal Ellen of X Y, therefore cannot combine this two terms together in tow. One natural law grow them. So let's now use both those properties and simplify the left hand side into one term, which will be Ellen of W to the zero point there. Eight times are 002 and the right hand side will say the same. So I'm just gonna put quotation work, so we have to write it. Okay, now let's go back to property of logarithms again. And look how we can get rid of this Ellen, because it's not in this top equation over here. So there's a property that's so naturally with natural algorithms. There's a property that says Ellen of X equals D. Therefore, e to the B equals X So very simply, we can put every we can raise, eat everything on the right hand side and make equal to the contents of this natural longer them. So let's do that. So E to the 0.2 are plus 0.0 one w plus C three, which is everything on the right hand side now equals the contents of the natural algorithm. Inside the brackets, a 0.8 times are 20.2 Okay, Now we're so close to getting the desired equation using rules of exponents, we know that we cannot separate this left hand side into three separate terms So e to the 0.0 to our times e to the 0.1 w times e to the C three equals this rate hand side. Now we know that if we simply divide by these 1st 2 terms on the left hand side, we can finally get, um, we can Finally I slate for a constant. However, we can mean also need to simply write c equals e to the C three were seized or final, arbitrary, constant in different books. We're gonna see different ways toe deal with the constant. Sometimes there's oversimplified. The best way is just to make sure that you keep your notation constant and right whenever you're changing if such as using c three or our final see. So now we finally get after dividing C equals don't bring you to the away are 0.2 over e to the 0.2 are then e to the 0.1 w and that is the final desired equation. Okay, now we need to grasp this, that I'm going through the 0.1000 40 so from figure three D that are is on the X axis, meaning that w is going to be on the y axis. So keep that in mind when we're graphing. So what we're gonna do is we need to know what this value see is when we graph. So in order to find that we simply need to plug in the values of R and W where are is 1000 ws 40 and plug into your calculator and you're gonna get a value that somewhere around 4.52 So, in order to enter this into your graphing calculator what you're going Teoh internal this original equation except C is going to become the 4.5 to value. And in your graphing calculator, you're going to get a shape that looks very similar to figure three like so, and it should be going through 1000 40. And keep in mind, however, that you see this bottom section in which W is less than zero. Since it's impossible to have a negative amount of wolves, you can simply ignore that

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Video Transcript

James Stewart

Calculus: Early Transcendentals

Catherine Ross

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Calculus: Early Transcendentals

Catherine Ross

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

James StewartCalculus: Early Transcendentals

Catherine Ross

Anna Marie Vagnozzi

Kristen Karbon

Samuel Hannah

James Stewart

Calculus: Early Transcendentals