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# Suppose we alter the differential equation in Exercise 23 as follows:$\frac {dP}{dt} = kP \cos^2 (rt - \phi) P(0) = P_0$(a) Solve this differential equation with the help of a table of integrals or a CAS.(b) Graph the solution for several values of $k, r,$ and $\phi.$ How do the values of $k, r,$ and $\phi$ affect the solution? What can you say about lim $_ {t \to \infty} P(t)$ in this case?

## a) $P=P_{0} \exp \left[\frac{k t}{2}+\frac{k}{4 r} \sin (2 r t-2 \phi)+\frac{k}{4 r} \sin (2 \phi)\right]$b) see graph

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Differential Equations

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##### Kristen K.

University of Michigan - Ann Arbor

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### Video Transcript

okay in this question. Were referring to a previous question which was the seasonal growth model here instead of just cosign, we have cosine squared. So let's write that down DP DT equals K. P. Yeah. Coulson squared rt minus phi with p of zero equals peanut. Now you could solve this by doing separation of variables. Uh but um says to use a table of integral Zohra cast. So what I did is I use mathematica. There's an are there a function called dissolve. And so I plugged in the differential equation with the initial condition. And this answer down here is the simplified version where that He is supposed to be the peanut. I also have s equals five because it is hard to put into mathematica but that's the equation. Then I graft a bunch of different versions of it. Okay, look here. So once again uh Yes. No that's on us. S mm equals uh thigh in the differential equation. So red corresponds to When the constant is five. Blue is when the constant is one and orange. I know this looks green. I don't have an orange option Is when the constant is equal to two. So in the first case I am changing K from 0.5 to 1 to two. So the bottom graph here is the red one. So that's when chemicals 10.5 and you can see that as K increases the direction that the function goes is much steeper. So it's going to go up more K. Is similar to a scaling factor larger K. Is the quicker the function is going to reach higher numbers. The next example is our so we see a bunch of different grass of different ours and they all seem kind of close. Uh and that shouldn't be too big of a surprise. That's because in the equation that we have, we have are in the numerator and denominator, so it's not going to play a huge role in that sense, but the R. Is in the sign was just determines how quickly it's going to alternate. So There is kind of like the period when R is equal 2.5, that's the red curve there can see that it's slowly changing from increasing to decreasing with those bumps. Um It's actually not decreasing, it just flattens out. But as you increase our there's going to be more of those changes. So it's going to that period of the sign is going to come by faster. So for example, the orange, there's a lot of those little bumps and then the last one is fi which I called S in this equation and that's pretty similar. It's more like a phase shift. So in the equation s doesn't play a huge role. It just determines when things are going to start. So all the curves are going to be pretty close but they might at different times go up and down. Okay. And last graph here is not necessary, but that's for different initial conditions. Okay, The last part asked limit as t goes to infinity of pft. So what happens is as t gets bigger and bigger, we know that dP DT It's always greater than or equal to zero because K is positive, cosine squared is At least zero and P is positive. So this is just going to keep on increasing. There's not gonna be a defined limit for it because it's going to go towards infinity. So PT approaches infinity and it shouldn't make sense because these graphs are kind of like exponential. To me, they look like wiggly exponential functions, an exponential blow up to infinity pretty quickly. So all these are going to blow up to infinity at different rates, but that is what's going to happen. They're going to look more and more like exponential and that's it.