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# (a) Show that the substitution $z = 1/P$ transforms the logistic differential equation $P' = kP(1 - P/M)$ into the linear differential equation$z' + kz = \frac {k}{M}$(b) Solve the linear differential equation in part (a) and thus obtain an expression for $P(t).$ Compare with Equation 9.4.7.

## a)$$z^{\prime}=-k z\left(1-\frac{1}{z M}\right) \Rightarrow z^{\prime}=-k z+\frac{k}{M} \Rightarrow z^{\prime}+k z=\frac{k}{M}(\star)$$b)$$P=\frac{M}{1+A e^{-k t}}$$

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Differential Equations

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Okay, Question. Given this logic differential equation Peters is k P Times one minus p by m And when we use a transformation, there equals to one by P We get a linear differential equation in terms off that and second part asking is to solve this differential equation and find barely off PT. Okay, so let's start with solution. So what is given to us? His speed ash K times P one minus B over em. And it is given you need to substitute status one hour p therefore peak and very tennis one hour. Is that okay? So different. Share this with respect Toe time, De so we get dp by. DT is one was a different station is minus one. Was that square times d said by DT But DP by DT is nothing but P dash and this is minus one. Was that quiet? Times that dash. Okay, so we have both the values to substitute. So we'll put this value here and we'll have peed ashes minus one. Who was that square times a dash? Is he calls to k Times P, which is one buys that say one minus p which is again one by said, And we have you in denominator m. Okay, so multiply by that square throughout, we'll get minus that dash. Is Kate times? That square was a test one minus. Zaid. Um, one was a dream. Okay, so you can cancel this wantem and when will multiply this zed inside So we'll have minus that dash is que Times zero minus one hour. I am. Okay, so now we need to remove this minus sign and write it in linear form. So first opened this bracket, so this becomes Case said minus K K by M. We'll ship this term in the right and this term and left so we'll have the dash. Plus, K said is equals to K by m, and this is a transformed equation. Now we can solve this way a method off linear differential equation. So comparing with y dash plus b y calls to queue R p value is K and Q values. Okay, bye. I am so are integrating factories. It is to integration off PDT. Here it is integration off ktt and since case constant will get it is to Katie. Okay, so we have our integrating factor. So answer is Zach Times it is to Katie equals to integration off by m times. It is to Katie DT and since gave I am is constant integration of various to Katie's. It is to Katie O K. Plus C. And this is that time she raised to Katie. Okay, so we can cancel this K. And so we get set as it is to Katie equals to one over m it is to Katie plus C Now, if we divided by it is to Katie, we get that is equals to one or m, since it is to Kate in is to get ticket cancel plus c times. It is two minus Katie. Since dividing, divided by rest located will give because we can keep. This is mind Isis rule one hour areas to m is areas too. My innocent. Okay, so we have our answer in terms off that. But we need answer in terms off Be so we know what s B from here we can see pes one over that, and therefore there will be one over Be okay, so we can substitute here. So 10 a p is equals to We can multiply this. So this becomes one plus M c. It is two minus Katie or am on inviting this we get P is equals to am upon one plus m times C times it is two minus. Katie. Now, this is a solution. But General, and we need particular solution. Then we need to find value off. See? Okay, but here we are not provided with condition. Toe find value of C so we can replace this with a new variable. A. So therefore p off t these em over one plus eight times. It is two minus Katie, and this is a final solution.

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Differential Equations

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