Write the solution of the initial value problem and use it to find the population when `t=20` ? `(dP)/(dt)=0.1P(1-P/2000) ` `P(0)=100` When does...

Sunday, December 25, 2011

Write the solution of the initial value problem and use it to find the population when $\displaystyle{t}={20}$ ? $\displaystyle\frac{{{d}{P}}}{{{\left.{d}{t}\right.}}}={0.1}{P}{\left({1}-\frac{{P}}{{2000}}\right)}$ $\displaystyle{P}{\left({0}\right)}={100}$ When does...

This is a separable differential equation. This means we can completely separate dependent and independent variables into two expressions. General form if such equation is

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{x}}}={f{{\left({x}\right)}}}{g{{\left({y}\right)}}}$

and the solution is obtained by solving the following integrals

$\displaystyle\int\frac{{\left.{d}{y}}}{{g{{\left({y}\right)}}}}=\int{f{{\left({x}\right)}}}{\left.{d}{x}\right.}+{C}$

Let us now return to the problem at hand.

$\displaystyle\frac{{\left.{d}{y}}}{{\left.{d}{t}}}={0.1}{P}{\left({1}-\frac{{P}}{{2000}}\right)}$

Now we need to put everything containing $\displaystyle{P}$ on the left and everything containing $\displaystyle{t}$ to the right side.

$\displaystyle\frac{{{d}{P}}}{{{0.1}{P}-\frac{{{P}}^{{2}}}{{20000}}}}={\left.{d}{t}\right.}$

Let us first simplify the expression on the left.

$\displaystyle\frac{{{d}{P}}}{{\frac{{{2000}{P}-{{P}}^{{2}}}}{{20000}}}}={\left.{d}{t}\right.}$

$\displaystyle\frac{{{20000}{d}{P}}}{{{P}{\left({2000}-{P}\right)}}}={\left.{d}{t}\right.}$

We shall write the term on the left using partial fractions to make integration easier.

$\displaystyle\frac{{{20000}{d}{P}}}{{{P}{\left({2000}-{P}\right)}}}=\frac{{A}}{{P}}+\frac{{B}}{{{2000}-{P}}}$

$\displaystyle{2000}{A}={20000}$

$\displaystyle{B}-{A}={0}$

$\displaystyle{A}={B}={10}$

$\displaystyle\frac{{{2000}{d}{P}}}{{P}}{\left({2000}-{P}\right)}=\frac{{10}}{{P}}+\frac{{10}}{{{2000}-{P}}}$

We can now integrate the equation.

$\displaystyle\int\frac{{10}}{{P}}{d}{P}+\int\frac{{10}}{{{2000}-{P}}}{d}{P}=\int{\left.{d}{t}\right.}$

$\displaystyle{10}{\ln{{P}}}-{10}{\ln{{\left({2000}-{P}\right)}}}+{10}{\ln{{C}}}={t}$

In the line above we have written the constant term as $\displaystyle{10}{\ln{{C}}}$ in stead of just $\displaystyle{C}.$ This is often used to make the expression easier to manipulate.

$\displaystyle{\ln{{P}}}-{\ln{{\left({2000}-{P}\right)}}}+{\ln{{C}}}=\frac{{t}}{{10}}$

Use formulae for logarithm of product and quotient:

$\displaystyle{\log}_{{a}}{\left({x}{y}\right)}={\log}_{{a}}{x}+{\log}_{{a}}{y}$

$\displaystyle{\log}_{{a}}{\left(\frac{{x}}{{y}}\right)}={\log}_{{a}}{x}-{\log}_{{a}}{y}$

$\displaystyle{\ln{{\left(\frac{{{C}{P}}}{{{2000}-{P}}}\right)}}}=\frac{{t}}{{10}}$

Take antilogarithm.

$\displaystyle\frac{{{C}{P}}}{{{2000}-{P}}}={{e}}^{{\frac{{t}}{{10}}}}$

$\displaystyle{C}{P}={2000}{{e}}^{{\frac{{t}}{{10}}}}-{{e}}^{{\frac{{t}}{{10}}}}{P}$

$\displaystyle{P}{\left({C}+{{e}}^{{\frac{{t}}{{10}}}}\right)}={2000}{{e}}^{{\frac{{t}}{{10}}}}$

$\displaystyle{P}=\frac{{{2000}{{e}}^{{\frac{{t}}{{10}}}}}}{{{C}+{{e}}^{{\frac{{t}}{{10}}}}}}$

We can now calculate $\displaystyle{C}$ by using the initial value.

$\displaystyle{P}{\left({0}\right)}={100}$

$\displaystyle\frac{{{2000}{{e}}^{{\frac{{0}}{{10}}}}}}{{{C}+{{e}}^{{\frac{{0}}{{10}}}}}}={100}$

Since $\displaystyle{{e}}^{{0}}={1},$ we have

$\displaystyle\frac{{2000}}{{{C}+{1}}}={100}$

$\displaystyle{C}+{1}=\frac{{2000}}{{100}}$

$\displaystyle{C}={20}-{1}$

$\displaystyle{C}={19}$

The solution to the initial value problem is

$\displaystyle{P}{\left({t}\right)}=\frac{{{2000}{{e}}^{{\frac{{t}}{{10}}}}}}{{{19}+{{e}}^{{\frac{{t}}{{10}}}}}}$

We can now calculate population when $\displaystyle{t}={20}.$

$\displaystyle{P}{\left({20}\right)}=\frac{{{2000}{{e}}^{{2}}}}{{{19}+{{e}}^{{2}}}}\approx{560}$

Population is approximately $\displaystyle{560}$ at time $\displaystyle{t}={20}.$

To find when the population reaches 1200, we need to solve the following equation

$\displaystyle\frac{{{2000}{{e}}^{{\frac{{t}}{{10}}}}}}{{{19}+{{e}}^{{\frac{{t}}{{10}}}}}}={1200}$

Multiply by the denominator.

$\displaystyle{2000}{{e}}^{{\frac{{t}}{{10}}}}={22800}+{1200}{{e}}^{{\frac{{t}}{{10}}}}$

$\displaystyle{800}{{e}}^{{\frac{{t}}{{10}}}}={22800}$

$\displaystyle{{e}}^{{\frac{{t}}{{10}}}}={28.5}$

Take logarithm.

$\displaystyle\frac{{t}}{{10}}={\ln{{28.5}}}$

$\displaystyle{t}={10}{\ln{{28.5}}}\approx{33.499}$

The population will reach 1200 at time $\displaystyle{t}={33.499}.$

Ninenotes

Sunday, December 25, 2011

Write the solution of the initial value problem and use it to find the population when $\displaystyle{t}={20}$ ? $\displaystyle\frac{{{d}{P}}}{{{\left.{d}{t}\right.}}}={0.1}{P}{\left({1}-\frac{{P}}{{2000}}\right)}$ $\displaystyle{P}{\left({0}\right)}={100}$ When does...

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find square roots of -1+2i

Sunday, December 25, 2011

Write the solution of the initial value problem and use it to find the population when ? When does...

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find square roots of -1+2i

Write the solution of the initial value problem and use it to find the population when $\displaystyle{t}={20}$ ? $\displaystyle\frac{{{d}{P}}}{{{\left.{d}{t}\right.}}}={0.1}{P}{\left({1}-\frac{{P}}{{2000}}\right)}$ $\displaystyle{P}{\left({0}\right)}={100}$ When does...