Chapter 7.4, Problem 22E

a.

To determine

the solution that satisfies the initial condition $y\left(0\right)=y_0$

Answer to Problem 22E

The solution that satisfies the initial condition $y\left(0\right)=y_0$ is $y={\left(-ckt+y_0^{-c}\right)}^{-\frac{1}{c}}$

Explanation of Solution

Given information : The differential equation is $\frac{dy}{dt}=k{y}^{1+c}$

Calculation : the solution of $y\left(0\right)=y_0$

  $\begin{array}{l}\frac{dy}{dt}=k{y}^{1+c}\\ {\displaystyle \int y^{-\left(1+c\right)}dy=k{\displaystyle \int dt}}\\ {\displaystyle \int y^{-c-1}dy=kt+C}\\ \frac{1}{-c}\times y^{-c}=kt+C\\ y^{-c}=-ckt+C\\ y={\left(-ckt+C\right)}^{-\frac{1}{c}}\\ y\left(0\right)={\left(-ck\left(0\right)+C\right)}^{{}^{--\frac{1}{c}}}\\ y\left(0\right)=C^{-\frac{1}{c}}\\ y_0^{-c}=C\\ y={\left(-ckt+y_0^{-c}\right)}^{-\frac{1}{c}}\end{array}$

b.

To determine

To show : that there is a finite time $t=T$ such that $\underset{t\to T-Y\left(t\right)}{\lim }=\infty$

Answer to Problem 22E

  $t=\frac{y_0^{-c}}{-ck}$

Explanation of Solution

Given information : The differential equation is $\frac{dy}{dt}=k{y}^{1+c}$

Calculation : from part (a)

  $\begin{array}{l}y={\left(-ckt+y_0^{-c}\right)}^{-\frac{1}{c}}\\ 0={\left(-ckt+y_0^{-c}\right)}^{-\frac{1}{c}}\\ 0=-ckt+y_0^{-c}\\ -ckt=-y_0^{-c}\\ t=\frac{y_0^{-c}}{-ck}\end{array}$

c.

To determine

To calculate : the growth of doomsday

Answer to Problem 22E

  $86.8579$ Months

Explanation of Solution

Given information : The question is

An especially prolific breed of rabbits has the growth term $k{y}^{101}$ . If two such as breed initially and the warren has $16$ rabbits after three month

Calculation :

  $\begin{array}{l}y={\left(-ckt+y_0^{-c}\right)}^{-\frac{1}{c}}\\ c=0.01\\ y_0=2\\ y\left(3\right)=16\\ 16={\left(-0.01\times k\times 3+2^{-0.01}\right)}^{-\frac{1}{0.01}}\\ {16}^{-0.01}=0.03k+2^{0.01}\\ {16}^{-0.01}-2^{0.01}=0.03k\\ k=\frac{{16}^{-0.01}-2^{0.01}}{-0.03}\approx 1.1434\\ y={\left(-ckt+y_0^{-c}\right)}^{-\frac{1}{c}}\\ y={\left(-0.01\times 1.1434\times t+2^{-0.01}\right)}^{-\frac{1}{0.01}}\\ 0=-0.01\times 1.1434\times t+2^{-0.01}\\ -2^{-0.01}=-0.01\times 1.1434\times t\\ t=\frac{-2^{-0.01}}{-0.01\times 1.1434}\approx 86.8579\end{array}$