Unformatted text preview: MEC702 Wk 3 Lect 2 Dierential Equations
• a Let y = f (x) be a function on real numbers. Then dierential equation is an equation which includes derivatives of y in it. • a rst order dierential equation • a second order dierential equation is a dierential equation of y
which includes the derviative of the highest order y (or
it. y is a dierential equation of
y 00 (or which includes the derivative of the highest order of
) in it.
dx2 And this goes on. Example.
1. Some examples include y 0 − y = e2x
(a) Example. or dy
dx − y = e2x . 2y 00 + y 0 − 3y = 4xy 2
Show that or 2 d y
2 + y = x3 + 3x + 1 dy
dx − 3y = −4xy 2 . satises y 000 + xy 00 − 2y 0 = 0. 1 Ordinary Dierential Equations (ODE)
• Ordinary Dierential Equation An (ODE) is an equation that contains one or several total derivatives of an unknown function y(t).
• For a function y = f (x), the total derivative is denoted as dy
and partial derivatives are denoted as ∂y
which is pronounced as del • A y over del Partial Dierential Equation x. (PDE) is an equation contain- ing partial derivatives of an unknown function of two or more
variables. • A general solution
stant c where c y(t) of an ODE contains an arbitrary con- can assume any real number. These produces innitely many solution curves. • A particular solution
for c. y(t) of an ODE has a particular value This represents one curve of the innitely many solution curves. 2 Initial Value Problems on y = f (x)
These are problems of these forms: dy
Remark. y(x0 ) = y0 Given an IVP, the procedure is the following: dy
dy = g(x)dx
y = G(x) + c
then use the initial value condition to nd the value of Exercise. c. Solve the following initial value problems (IVP). 1. dy
dx 2. dy
dx 3. dy
dx = cos x, y(0) = 1.
= sin t + 1, y π3 =
= 3 x, y(1) = 2 4. dy
dx = x+1
2 y(1) = 0 3 Exponential Growth and Decay
This modelling is a type of Initial Value Problem with applications
on population growth, spread of diseases, medicine, radioactive decay, radiocarbon dating, etc. Population Growth
One of the simplest models of population growth is based on the
observation that when populations (people, bacteria, plants, fruit
ies, etc) are not constrained by environmental limitations, they
tend to grow at a rate proportional to the size of the population.
That is, the larger the population, the more rapidly it grows.
rate of ∝ population growth
over time population dy
Hence, we have the size of ∝ y dierential equation for population growth
dt where k is a positive proportionality constant. 4 Pharmacology
When a drug (say, panadol) is administered to an individual, it
enters the bloodstream and then is absorbed by the body over time.
Medical research has shown that the amount of drug that is present
in the bloodstream tends to decrease at a rate that is proportional
to the amount of drug present - the more of the drug that is present
in the bloodstream, the more rapidly it is absorbed by the body.
absorption amount of ∝ drug present over time in the body
to be absorbed dy
dt Hence, we have the ∝ y dierential equation for drug absorption
dt where k is a positive proportionality constant. The negative sign is required because y decreases with time. Radioactive Decay
It is a fact of Physics that radioactive elements disintegrate sponta- radioactive decay . neously in a process called Radioactive elements are those elements which lose matter over time. This may be due
to the unstability of the bonding between nucleus and electrons.
Hence, they literally give o energy and matter in all directions.
But as time goes on, the amount of matter left decreases, hence
decreases the rate of decay of the material.
decay amount of ∝ material left over time dy
dt Hence, we have the to be decayed ∝ y dierential equation for radioactive decay
5 Exponential Growth and Decay Model
Given the dierential equation, we nd the function y(t): dy
Since c is a constant, = kdt
= (kt + c)
= e(kt+c) = ekt · ec
= Aekt . A = ec is just gives another constant. Hence, the function representing exponential growth is y(t) = Aekt .
Similarly, the function representing exponential decay is y(t) = Ae−kt .
Sometimes, there is an initial condition y(0) = c given, which comes to y(0) = Ae0 = A. 6 Example.
in 1998 According to United Nations data, the world population
was approximately 5.9 billion (5.9 × 10 ) and growing at a rate of about 1.33% per year. Assuming an exponential growth model, estimate the world population at the beginning of the year 2023. What about the population at the beginning of Example.
20 A cell of the bacterium E. coli divides into two cells every minutes when placed in a nutrient culture. the number of cells that are present
placed in the culture. t 2015? Let y = y(t) be minutes after a single cell is Assume that the growth of the bacteria is approximated by a continuous exponential growth model.
1. Find an initial value problem whose solution is
2. Find a formula for y(t). y(t). 3. How many cells are present after 2 hours? 4. How long does it take for the number of cells to reach Example. 1 000 000? A scientist wants to determine the half-life of a certain radioactive substance. She determines that in exactly
milligram sample of a substance decays to 5 days, a 10.0 3.5 milligrams. Based on the data, what is the half-life? [Half-life is the time it takes for a
substance to decay to half its initial size] 7 ...
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