$\begin{array}{1 1} 97600 \\ 197600 \\ 234600 \\ 234419 \end{array} $

- First order,first degree DE variable separatable : Variable of a DE are rearranged to separate there (i.e.) $f_1(x)g_2(y)dx+f_2(x)g_1(y)dy=0$
- It can be written as $\large\frac{g_1(y)}{g_2(y)}$$dy=\large\frac{f_1(x)}{f_2(x)}$$dx$
- The solution is therefore $\int \large\frac{g_1(y)}{g_2(y)}$$dy=-\int \large\frac{f_1(x)}{f_2(x)}$$dx+c$

Step 1:

Let $x$ be the population of the city at time $t$ in years

$t=0$ in the year 1960,when the initial population $x_0=130,000$

Step 2:

Now $\large\frac{dx}{dt}\propto x$

$\therefore \large\frac{dx}{dt}$$=kx$

Step 3:

$\int \large\frac{dx}{x}$$=\int kdt+\log c$

$\log x=kt+\log c$

$\log \large\frac{x}{c}=$$kt$

$x=ce^{kt}$

Step 4:

When $t=0,x=1,30,000$

$\therefore c=1,30,000$

$\Rightarrow x=130000e^{kt}$

Step 5:

In 1990,t=30 years and $x=16,000$

$\therefore 1,60,000=1,30,000e^{30k}=\large\frac{16}{13}$

$30k\log e=\log \large\frac{16}{13}$

$\log_e\large\frac{16}{13}$$=0.2070$

$k=\large\frac{0.2070}{30}$

$k=0.0069$

Step 6:

In 2020,$t=60years$

$x=130000e^{60k}$

$\;\;=130000e^{0.0069\times 60}$

$\;\;\;=130000e^{0.42}$

$e^{0.42}=1.52$

$\;\;\;=130000\times 1.52$

$\;\;\;=197600$

The population in 2020 is anticipated to be 197600

Ask Question

Tag:MathPhyChemBioOther

Take Test

...