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Firlakuza [10]
3 years ago
5

Town A adds 10 people per year to its population, and town B grows by 10% each year. In 2006, each town has 145 residents. For e

ach town, determine whether the population growth is linear or exponential. Match the town on the left with the description of its population growth on the right.
Mathematics
1 answer:
avanturin [10]3 years ago
7 0
Well town A has an annual growth that stays constant. town B has a 10% growth which is exponential because it takes a percent of the whole number and adds it so if you have 145 residents in 2006 in say three years in 2009. in 2007 its population would be 159.5> rounded up because you can't have half a person it would be 160. in 2008 it would be 176 and finally in 2009 it would be 194. that's a growth of 49 in three years where as town a grows only ten people per year. which means town a would only grow by 30 people in three years and have a population of 175.
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y' 5y = e^{4t}, y(0) = 2 is  mathematically given as

y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}

<h3>What is the Laplace transform of the given initial-value problem? y' 5y = e4t, y(0) = 2?</h3>

Generally, the equation for the problem is  mathematically given as

&\text { Sol:- } \quad y^{\prime}+s y=e^{4 t}, y(0)=2 \\\\&\text { Taking Laplace transform of (1) } \\\\&\quad L\left[y^{\prime}+5 y\right]=\left[\left[e^{4 t}\right]\right. \\\\&\Rightarrow \quad L\left[y^{\prime}\right]+5 L[y]=\frac{1}{s-4} \\\\&\Rightarrow \quad s y(s)-y(0)+5 y(s)=\frac{1}{s-4} \\\\&\Rightarrow \quad(s+5) y(s)=\frac{1}{s-4}+2 \\\\&\Rightarrow \quad y(s)=\frac{1}{s+5}\left[\frac{1}{s-4}+2\right]=\frac{2 s-7}{(s+5)(s-4)}\end{aligned}

\begin{aligned}&\text { Let } \frac{2 s-7}{(s+5)(s-4)}=\frac{a_{0}}{s-4}+\frac{a_{1}}{s+5} \\&\Rightarrow 2 s-7=a_{0}(s+s)+a_{1}(s-4)\end{aligned}

put $s=-s \Rightarrow a_{1}=\frac{17}{9}$

\begin{aligned}\text { put } s &=4 \Rightarrow a_{0}=\frac{1}{9} \\\Rightarrow \quad y(s) &=\frac{1}{9(s-4)}+\frac{17}{9(s+s)}\end{aligned}

In conclusion, Taking inverse Laplace tranoform

L^{-1}[y(s)]=\frac{1}{9} L^{-1}\left[\frac{1}{s-4}\right]+\frac{17}{9} L^{-1}\left[\frac{1}{s+5}\right]$ \\\\

y(t)=\frac{1}{9} e^{4 t}+\frac{17}{9} e^{-5 t}

Read more about Laplace tranoform

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Step-by-step explanation:

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