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Chapter day 4 Differential equations

Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

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Page 1: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

Chapter day 4Differential equations

Page 2: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present (at least for awhile.)

So does any population of living creatures. Other things that increase or decrease at a rate proportional to the amount present include radioactive material and money in an interest-bearing account.

If the rate of change is proportional to the amount present, the change can be modeled by:

dyky

dt

Recall from AP Calculus

Page 3: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

dyky

dt

1 dy k dt

y

1 dy k dt

y

ln y kt C

Rate of change is proportional to the amount present.

Divide both sides by y.

Integrate both sides.

Page 4: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

1 dy k dt

y

ln y kt C

Integrate both sides.

Exponentiate both sides.

When multiplying like bases, add exponents. So added exponents can be written as multiplication.

ln y kt Ce e

C kty e e

Page 5: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

ln y kt Ce e

C kty e e

Exponentiate both sides.

When multiplying like bases, add exponents. So added exponents can be written as multiplication.

C kty e e

kty Ae Since is a constant, let .Ce Ce A

Page 6: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

C kty e e

kty Ae Since is a constant, let .Ce Ce A

At , .0t 0y y00

ky Ae

0y A

1

0kty y e This is the solution to our original initial

value problem.

Page 7: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

0kty y eWe end with:

dyky

dtSo if we start with:

Page 8: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

What if we have a series of differential equations?

dy1 = ky1

dt

dy2 = ky2

dt

dy3 = ky3

dt

We could solve of these individually

y1 =c1ekt

y2 =c2ekt

y3 =c3ekt

Provided that we have initial conditions for each of these to solve for the constants

Page 9: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

If we define x

x1’(t) x’(t) = x2’(t) xn ’(t)

[ ]…

This yields the equation x’(t)= AxWhich is easy to solve in the case of a diagonal matrix.

x’1x’2x’3

[ ] = 3 0 00 -2 00 0 4[ ] x1

x2

x3[ ]

We can solve each of these as a separate differential equationx1’ = 3x1, x2’ = -2x2, x3’ = 4x3

x1 (t) = b1e3t, x2 (t) = b2e-2t, x3 (t) = b3e4t, This is the general solution. We can solve for the constants if given an initial condition.

Page 10: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

First order homogeneous linear system of differential equations

x1’(t) = a11x1 (t) + a12 x2 (t) + … a1nxn (t)

x2’(t) = a21x1 (t) + a22 x2 (t) + … a2nxn (t)

xn ’(t) = an1x1 (t) + an2 x2 (t) + … annxn (t)

…We could write this in matrix form as: x1’ (t) a11 a12 …. a12

x(t) = x2’ (t) A = a21 a 22 … a2n

xn’ (t) an1 a n2 …anm

[ ]… [ ]…

Page 11: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

What if our system is not diagonal?

du1 = -u1 +2 u2

dt

du2 = u1 – 2u2

dt

A = -1 2 1 -2 [ ]

The system at the left can be written as du/dt = Au with a as

How can we solve this system?Initial condition u(0) =

10[ ]

Page 12: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

du/dt =Au

y= eAt

u(t) = c1eλ t x1 +c2e λ t x2+…+ cneλ t

xn

Check that each piece solves the given system

du/dt =Au

d (eλ t x1) = A eλ t

x1 λeλ t x1 = A eλ t

x1

dt λx1 =Ax1

1 2 n

Page 13: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

Key Formulas

Difference Equations

Differential Equationsdu/dt =Au y= eAt

Page 14: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

Solve the differential equations

A = -1 2 1 -2 [ ]The system at the left can be written as du/dt = Au with a as

What are the eigenvalues from inspection?Hint: A is singularThe trace is -3

Start by computing the eigenvalues and eigenvectors

Page 15: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

Solve the differential equationsStep 1 find the eigenvalues and

eigenvectors

det -1-λ 2 1 -2-λ [ ]

We can a solve via finding the determinant of A - λI

Calculate the eigenvector associated with λ = 0,-3

By inspection: the matrix is singular therefore 0 is an eigenvalue the trace is -3 therefore the other eigenvalue is -3

A = -1 2 1 -2[ ] For λ = 0 find a basis for the kernel of A

21[ ]

For λ= -3 find a basis for the kernel of A+3IA+ 3I = 2 2 1 1[ ] 1

-1[ ]

Page 16: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

Solve the differential equations

A = -1 2 1 -2 [ ]

The system at the left can be written as du/dt = Au with a as

The form that we are expecting for the answer is y = c1 e λ t x1 + c2 e λ t x2

Note: the solutions of the equations are going to be e raised to a power.

1 2

The eigenvalues are already telling us about the form of the solutionsA negative eigenvalue will mean that that portion goes to zero as x goes to infinity. An eigenvalue of zero will mean that we will have an e0 which will be a constant. We will call this type of system a steady state.

Page 17: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

Solve the differential equations

A = -1 2 1 -2 [ ]

Solve by plugging in eigenvalues into expected equation and for λ1 and λ2. and the corresponding eigenvectors in x1 and x2

We find c1 and c2 by using the initial condition

y = c1 e0t 2 + c2 e -3t 1 1 -1[ ] [ ]

Initial condition u(0) = []10

Plugging in zero for t and the initial conditions yields:

1 = c1 2 + c2 10 1 -1[ ][ ][]

Recall:

c1 = 1/3c2 = 1/3

Page 18: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

Solve the differential equations

The general solution isy = 1/3 2 + 1/3 e -3t 1 1 -1[ ] [ ]

We are interested in hat happens as time goes to infinityRecall our initial condition was 1 all of our quantity was in u1

0

Then as time progressed there was flow from u1 to u2. As time approaches infinity we end with the steady state 2/3 1/3[ ]

[]

Page 19: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

The solution to y’ = ky is y = y0ekt

The solution to x’ = Au

is u = c0eAt

Page 20: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

Applications of Linear Algebra

• Flow of water, electricity or money through a net work that continues over time.

Page 21: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

Applications of Differential Equations

• Differential Equations are the language in which the laws of nature are expressed. Understanding properties of solutions of differential equations is fundamental to much of contemporary science and engineering. Ordinary differential equations (ODE's) deal with functions of one variable, which can often be thought of as time

- MIT

Page 22: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

http://en.wikipedia.org/wiki/Differential_equation

• Many fundamental laws of physics and chemistry can be formulated as differential equations. In biology and economics, differential equations are used to model the behavior of complex systems. The mathematical theory of differential equations first developed together with the sciences where the equations had originated and where the results found application. However, diverse problems, sometimes originating in quite distinct scientific fields, may give rise to identical differential equations. Whenever this happens, mathematical theory behind the equations can be viewed as a unifying principle behind diverse phenomena. As an example, consider propagation of light and sound in the atmosphere, and of waves on the surface of a pond. All of them may be described by the same second-order partial differential equation, the wave equation, which allows us to think of light and sound as forms of waves, much like familiar waves in the water. Conduction of heat, the theory of which was developed by Joseph Fourier, is governed by another second-order partial differential equation, the heat equation. It turned out that many diffusion processes, while seemingly different, are described by the same equation; the Black–Scholes equation in finance is, for instance, related to the heat equation.

Page 23: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

All of the following are stated in terms of differential equations

• Newton’s second law of dynamics• Euler –Lagrange theorem Classical mechanics• Radioactive decay – nuclear physics• Newton’s Law of cooling – Thermodynamics• Maxwell’s equation – electro magnetism• Einstein’s field equation – General relativity• The Shroedinger equation – Quantum mechanics• Mathusian growth model – Economics• Verhulst equation – biological population growth• http://en.wikipedia.org/wiki/Differential_equation

Page 24: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

• http://ocw.mit.edu/courses/mathematics/18-03-

differential-equations-spring-2010/

Page 26: Chapter day 4 Differential equations. The number of rabbits in a population increases at a rate that is proportional to the number of rabbits present

What if the matrix is not diagonal?

• White book p. 520 ex 3, 4, 5