Transcript Day-4-differential-equations

Day 4 Differential Equations
(option chapter)
Recall from AP Calculus
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:
dy
 ky
dt

dy
 ky
dt
1 dy
=k
y dt
1
 y dy   k dt
Rate of change is proportional
to the amount present.
Divide both sides by y.
Integrate both sides.
ln y = k t +C

1
 y dy   k dt
Integrate both sides.
ln y = k t +C
e
ln y
=e
kt + C
y = e ×e
C
kt
Exponentiate both sides.
When multiplying like bases, add
exponents. So added exponents
can be written as multiplication.

ln y
e
e
Exponentiate both sides.
kt C
y  e e
C
When multiplying like bases, add
exponents. So added exponents
can be written as multiplication.
kt
y  e e
C kt
y  Ae
kt
Since
 eC
is a constant, let  e
C
A.

y  e e
C kt
y  Ae
kt
y0  Ae
Since
 eC
is a constant, let  e
C
A.
1
k 0
At
t  0 , y  y0
.
y0  A
y  y0 e
kt
This is the solution to our original initial
value problem.

So if we start with:
We end with:
dy
 ky
dt
y  y0 e
kt

What if we have a series of
differential equations?
dy1
= ky1
dt
dy2
= ky2
dt
dy3
= ky3
dt
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
x’(t) =
[ ]
…
x1’(t)
x2’(t)
xn ’(t)
This yields the equation x’(t)= Ax
Which is easy to solve in the case of a
diagonal matrix.
[ ] [ ][ ]
x’1
x’2
x’3
=
3 0 0
0 -2 0
0 0 4
x1
x2
x3
We can solve each of these as a separate differential equation
x1’ = 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?
The system at the left can be written
as du/dt = Au with a as
du1 = -u1 + 2u2
dt
[ ]
A = -1 2
1 -2
du2 = u1 – 2u2
dt
[]
Initial condition u(0) =
1
0
How can we solve this system?
du/dt = Au
y = eAt
u(t) = c1eλ1 t x1 +c2e λ2 t x2+…+ cneλ n t xn
Check that each piece solves the given system
du/dt = Au
d (eλ t x1) = A eλ t x1
dt
λeλ t x1 = A eλ t x1
λx1 = Ax1
Key Formulas
Difference Equations
Differential Equations
du/dt = Au
y = eAt
Solve the differential equations
The system at the left can be written
as du/dt = Au with a as
[ ]
A = -1 2
1 -2
Start by computing the
eigenvalues and eigenvectors
What are the eigenvalues from inspection?
Hint: A is singular
The trace is -3
Solve the differential equations
Step 1 find the eigenvalues and
eigenvectors
We can a solve via finding the
determinant of A - λI
By inspection: the matrix is singular
det -1-λ 2
therefore 0 is an eigenvalue the trace is
1 -2-λ
-3 therefore the other eigenvalue is -3.
[ ]
Calculate the eigenvector associated with λ = 0, -3
[ ]
[ ]
A = -1 2
1 -2
A+ 3I = 2
1
2
1
For λ = 0 find a basis for the kernel of A
[]
[]
2
1
For λ= -3 find a basis for the kernel of A+3I
1
-1
Solve the differential equations
The system at the left can be written
as du/dt = Au with a as
Note: the solutions of the equations
are going to be e raised to a
power.
[ ]
A = -1 2
1 -2
The form that we are expecting for the answer is
y = c1 e λ t x1 + c2 e λ t x2
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
Solve by plugging in eigenvalues into expected equation and for λ1 and
λ2. and the corresponding eigenvectors in x1 and x2
[ ]
A = -1 2
1 -2
y = c1 e0t 2 + c2 e -3t 1
1
-1
[]
We find c1 and c2 by
using the initial condition
Recall:
Initial condition u(0) = 1
0
c1 = 1/3
c2 = 1/3
[]
Plugging in zero for t
and the initial conditions yields:
[] [] [ ] [ ]
1
0
= c1
2
1
+ c2 1
-1
Solve the differential equations
The general solution is
y = 1/3
2 + 1/3 e -3t
1
[]
[]
1
-1
We are interested in hat happens as time goes to infinity
Recall 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