MEC702_Wk_3_Lect_2 - MEC702 Wk 3 Lect 2 Dierential...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MEC702 Wk 3 Lect 2 Dierential Equations Denition. • 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 dy 0 which includes the derviative of the highest order y (or ) in dx it. y is a dierential equation of y 00 (or which includes the derivative of the highest order of d2 y ) 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 dx 2 + y = x3 + 3x + 1 dy dx − 3y = −4xy 2 . satises y 000 + xy 00 − 2y 0 = 0. 1 Ordinary Dierential Equations (ODE) Denition. • 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 dx and partial derivatives are denoted as ∂y ∂x 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 = g(x), dx Remark. y(x0 ) = y0 Given an IVP, the procedure is the following: dy = g(x) dx dy = g(x)dx ˆ ˆ 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 √ , 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 dt Hence, we have the size of ∝ y dierential equation for population growth dy = ky 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. rate of 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 dy = −ky 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. rate of decay amount of ∝ material left over time dy dt Hence, we have the to be decayed ∝ y dierential equation for radioactive decay dy = −ky. dt 5 Exponential Growth and Decay Model Given the dierential equation, we nd the function y(t): dy = ky dt 1 dy y ˆ 1 dy y ln y y y Since c is a constant, = kdt ˆ = 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 9 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 ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern