Day 4 Differential Equations(option chapter)

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

dyky

dt

1 dy k dt

y

Rate of change is proportional to the amount present.

Divide both sides by y.

Integrate both sides.

1 dy k dt

y 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

Exponentiate both sides.

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

C kty e ekty Ae Since is a constant, let .

Ce Ce A

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.

0kty y eWe end with:

dyky

dtSo if we start with:

What if we have a series of differential equations?

We could solve these individually

y1 =c1ekt

y2 =c2ekt

y3 =c3ekt

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

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.

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

[ ]… [ ]…

What if our system is not diagonal?

du1 = -u1 + 2u2

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[ ]

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

Key Formulas

Difference Equations

Differential Equationsdu/dt = Au y = eAt

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

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[ ]

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 solutions.A 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.

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

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[ ]

[]

The solution to y’ = ky is y = y0ekt

The solution to x’ = Au

is u = c0eAt

Homework: wkst 8.4 1-9 odd, 2 and 8

What if the matrix is not diagonal?

• White book p. 520 ex 3, 4, 5