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course Mth 279
22 July 13, 10:11pm
Query 16 Differential Equations*********************************************
Question: Find the general solution to
y '' - 5 y ' + 2 y = 0
and the unique solution for the initial conditions y (0) = -1, y ' (0) = -5.
How does the solution behave as t -> infinity, and as t -> -infinity>?
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Your solution:
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y '' - 5 y ' + 2 y = 0
let y(t) = A*e^(rt)
then y'(t) = r*A*e^(rt)
and y""(t) = r^2*A*e^(rt)
r^2*A*e^(rt) - 5*r*A*e^(rt) + 2*A*e^(rt) = 0
r^2 - 5r + 2 = 0 (quadratic equation)
r = (5 +/- sqrt(17))/2
r = 4.54, 0.438
so,
y = Ae^(4.56t) + Be^(0.438t)
use initial conditions to solve for A and B
at y(0) = -1 then A + B = -1
at y'(o) = -5 then 4.56A + o.438B = -5
Solving simultaneously, A = -5.047 and B = 0.004675
y = -5.05e^(4.56t) + 0.00468e^(0.438t)
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Very clearly your solution for t = 0 does not equal -1.
I got approximate solutions A = `1.1, b = 0.1.
As t -> infinity the exponents of both both terms remain positive and become large, with the negative term having the greater magnitude; as a result the limit is -infinity.
As t -> - infinity both exponents approach -infinity so both terms approach zero, resulting in a limit of zero.
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Question: Find the general solution to
8 y '' - 6 y ' + y = 0
and the unique solution for the initial conditions y (1) = 4, y ' (1) = 3/2.
How does the solution behave as t -> infinity, and as t -> -infinity>?
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Your solution:
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8 y"" - 6 y' + y = 0
y"" - (3/4)y' + (1/8)y = 0
Use y = Ae^(rt)
solve for y' and y""
plug y, y',and y"" into equation
r^2 - (3/4)r + (1/8)r = 0
solve for r using quadratic formula
r = ((3/4) +/- sqrt((3/4^2 - 1/2)))/2
r = 0.406, 0.344
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Decimal approximations are sometimes appropriate but exact solutions have no roundoff error and are both accurate and safe.
The characteristic equation is
8 r^2 - 6 r + 1 = 0
with solutions
r = (6 +- sqrt(36 - 4 * 8 * 1) ) / (2 * 8) = (6 +- sqrt(4)) / 8 = (6 +- 2) / 16
so solutions are
r = 8/16 = 1/2
and
r = 4/16 = 1/4.
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y = Ae^(0.406t) + Be^(0.344t)
y' = 0.406*Ae^(0.406t) + 0.344Be^(0.344t)
Solve for A and B at initial condition point
y(1) = 4 = 0.406Ae^0.406 + 0.344Be^0.344
y'(1) = 3/2 = 0.406Ae^0.406 + 3.44Be^0.344
solve simultaneously
A = 1.33 and B = 1.42
So, y = 1.33e^0.406t + 1.42e^0.344t
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Question: Solve the equation
m ( r '' - Omega^2 r) = - k r ' for r(0) = 0, r ' (0) = v_0.
Omega is the angular velocity of a centrifuge, m is the mass of a particle and k is drag force constant. Physics students will recognize that m r Omega^2 is
the centripetal force required to keep an object of mass m moving in a circle of radius r at angular velocity Omega.
The equation models the motion of a particle at the axis which is given initial radial velocity v_0.
The mass m of a particle is proportional to its volume, while the drag constant is proportional to its cross-sectional area. Assuming all particles are geometrically
similar (and most likely spherical, though this is not necessary as long as they are geometrically similar, but for the sake of an accurate experiment spheres would be
preferable), how then does k / m change as the diameter of particles increases?
If Omega = 20 revolutions / minute and v_0 = 1 cm / second, with k / m = 4 s^-1, find r( 2 seconds ).
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Your solution:
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m (r"" - Omega^2 r) = - k r'
solve to put in standard form
r"" + (k/m)r - Omega^2 r = 0
plug in values given in problem
r"" + 4r' - 20^2 r = 0
let r = Ae^(kt)
solve for r' and r""
plug r, r', and r"" into equation and divide out Ae^(kt)
k^2 + 4k - 400 = 0
solve for k using quadratic equation
k = 18.1, -22.1
r(t) = Ae^(18.1t) + Be^(-22.1t)
r'(t) = 18.1Ae^(18.1t) - 22.1Be^(-22.1t)
solve for A and B at given initial conditions
A = 0.0249
B = -0.0249
r = 0.0249e^(18.1t) - 0.0249e^(-22.1t)
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Your constants need to have compatible units. rev / minute doesn't articulate with SI units.
The equation yields general solution
r(t) = A e^( (-k / (2 m) ( 1 + sqrt( 1 + 4 m^2 omega^2 / k^2) ) * t ) + B e^( (-k / (2 m) ( 1 - sqrt( 1 + 4 m^2 omega^2 / k^2) ) * t ).
Now given that k / m = 4 s^-1 and omega = 20 rev / min = 2 pi / 3 rad / sec we can put some numbers into our solution.
We evaluate the factor 1 + 4 m^2 omega^2 / k^2:
1 + 4 m^2 omega^2 / k^2 =
1 + 4 (m / k)^2 * omega^2 =
1 + 4 omega^2 / (k/m)^2 =
1 + 4 (2 pi / 3 s^-1)^2 / (4 s^-1)^2 =
1 + pi^2 / 9,
so that
sqrt( 1 + 4 m^2 omega^2 / k^2) = sqrt( 1 + pi^2 / 9) = 1.45, approx..
Our function becomes approximately
r(t) = A e^( ( -2 s^-1) ( 1 + 1.45) t) + B e^(-2 s^-1) ( 1 - 1.45) t) = A e^(-4.9 t) + B e^(.9 t)
with velocity function
v(t) = r ' (t) = -4.9 A e^(-4.9 t) + .9 B e^(.9 t).
r(0) = 0 and v(0) = r ' (0) = 1 cm/s so
A + B = 0
-4.9 A + .9 B = 1
This yields approximate solution A = -.17, B = .17, both in units of cm.
Thus
r(t) = -.17 e^(-4.9 t) + .17 e^(.9 t)
and
r(2) = 1.03 cm.
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Good approach overall, but with some errors in details.
Check my notes.
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