Notes: Write and solve the differential equation that models the verbal statement. Evaluate the solution at the specified value of the independent...

Thursday, 23 January 2014

Write and solve the differential equation that models the verbal statement. Evaluate the solution at the specified value of the independent...

The rate of change of N is the derivative of N with respect to t, or $\displaystyle\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}$ . If the rate of change of N is proportional to N, then

$\displaystyle\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}={k}{N}$ , where k is the proportionality constant. This is the differential equation we need to solve.

To solve it, separate the variables:

$\displaystyle\frac{{{d}{N}}}{{N}}={k}{\left.{d}{t}\right.}$

Integrating both sides results in

$\displaystyle{\ln{{N}}}={k}{t}+{C}$ , where C is another constant. This can...

The rate of change of N is the derivative of N with respect to t, or $\displaystyle\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}$ . If the rate of change of N is proportional to N, then

$\displaystyle\frac{{{d}{N}}}{{{\left.{d}{t}\right.}}}={k}{N}$ , where k is the proportionality constant. This is the differential equation we need to solve.

To solve it, separate the variables:

$\displaystyle\frac{{{d}{N}}}{{N}}={k}{\left.{d}{t}\right.}$

Integrating both sides results in

$\displaystyle{\ln{{N}}}={k}{t}+{C}$ , where C is another constant. This can be rewritten in exponential form as

$\displaystyle{N}={{e}}^{{{k}{t}+{C}}}={N}_{{0}}{{e}}^{{{k}{t}}}$ . Here, $\displaystyle{N}_{{0}}={{e}}^{{C}}$ and it equals N(t) when t = 0.

When t = 0, N = 250, so

$\displaystyle{N}{\left({0}\right)}={N}_{{0}}={250}$ and $\displaystyle{N}{\left({t}\right)}={250}{{e}}^{{{k}{t}}}$ is the solution of the differential equation above with the initial condition N(0) = 250.