#$&*
course Mth 277
4-10 11
Query 16 Differential Equations*********************************************
Question: Find the general solution to
y '' - 5 y ' + 2 y = 0
and the unique solution for the initial conditions y (0) = -1, y ' (0) = -5.
How does the solution behave as t -> infinity, and as t -> -infinity>?
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Your solution:
Characteristic polynomial for y '' - 5 y ' + 2 y = 0 is,
P(`lamda) = `lambda^2 - 5(`lambda) + 2
`lambda = [-(-5) +- `sqrt((-5)^2 - 4*2)]/2
= [5 +- `sqrt(17)]/2 = 2.5 +- `sqrt(17)/2
So we have….
y1(t) = e^((2.5 + `sqrt(17)/2) t) and y2(t) = e^((2.5 - `sqrt(17)/2) t)
General solution is
y(t) = c1e^((2.5 + `sqrt(17)/2) t) + c2e^((2.5 - `sqrt(17)/2) t)
Unique solution we must evaluate derivatives
y’(t) =(2.5 + `sqrt(17)/2*c1e^((2.5 + `sqrt(17)/2) t) + (2.5 - `sqrt(17)/2*c2e^((2.5 - `sqrt(17)/2) t)
So..
c1 + c2 = -1
(2.5 + `sqrt(17)/2*c1 + (2.5 - `sqrt(17)/2*c2 = -5
Mult the first equation by -(2.5 + `sqrt(17)/2 we have new system
-(2.5 + `sqrt(17)/2c1 - (2.5 + `sqrt(17)/2c2 = (2.5 + `sqrt(17)/2
(2.5 + `sqrt(17)/2*c1 + (2.5 - `sqrt(17)/2*c2 = -5
Solving the following
(2.5 - `sqrt(17)/2*c2 - (2.5 + `sqrt(17)/2c2 = -5 + (2.5 + `sqrt(17)/2
c2 = approx. .1063
c1= approx. -1.1063
As t infinity because both exponents are pos the solution approaches pos infintity as t -infinity solution approaches 0.
confidence rating #$&*:
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Given Solution:
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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>?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
Characteristic polynomial for 8 y '' - 6 y ' + y = 0 is,
P(`lamda) = 8(`lambda^2) - 6(`lambda) + 1
`lambda = .25 and .5
So we have….
y1(t) = e^(.25t) and y2(t) = e^(.5t)
General solution is
y(t) = c1e^(.25t) + c2e^(.5t)
Unique solution we must evaluate derivatives
y’(t) =.25c1e^(.25t) + .5c2e^(.5t)
So..
c1 + c2 = 4
.25c1 + .5c2 = 3/2
c2 = 2
c1= 2
As t infinity because both exponents are pos the solution approaches pos infinity as t -infinity solution approaches 0.
@& Your solution is correct if the initial conditions are y(0) = 4 and y ' (0) = 3/2.
If the initial conditions as given here are correct then we get approximate solutions
c_1 = 1.60
c_2 = 1.06
with the corresponding general solution.
However I would be that the initial conditions are the ones you solved for. *@
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Given Solution:
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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:
As the diameter increases both volume and c.s. area increase but volume will increase at a faster rate. Where mass is proportional to volume and and drag constant is proportaional to c.s. area k/m will get smaller and smaller as diameter increases.
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Given Solution:
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@& Check the following solution and see if you concur:
Using trial solution e^(lambda t) our characteristic equation will be
lambda^2+ k/m lambda - omega^2 = 0
with solutions
lambda = (-k / (2 m) +- sqrt(k^2 / m^2 + 4 omega^2) / 2 ) = -k / (2m) * ( 1 +- sqrt( 1 + 4 m^2 omega^2 / k^2) )
yielding 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.
Note that 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.
*@
@&
Very good work, but be sure to see my notes, especially on that last problem.*@