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course Mth 279
5/5
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:
Using a certain method given in the book required the Quadratic formula for:
substitute a = Lanbda
a^2 -5a + 2 = 0
a_1,2 = (-b +,- sqrt(b^2 -4ac))/(2a) = (5 +,- sqrt(25 - 4(1)(2)))/(2(1))
a1= (5 + sqrt(17))/2 = 4.562 approx.
or
a2= (5 - sqrt(17))/2 = 0.438 approx.
y(t) = Ae^((5 + sqrt(17))t/2) + Be^((5 - sqrt(17))t/2)
when y(0)
y(0) = Ae^0 + Be^0 = -1
A + B = -1
y'(0) = -5
y'(t) = A((5+sqrt(17))/2)e^((5 + sqrt(17))t/2) + B((5-sqrt(17))/2)e^((5 - sqrt(17))t/2)
y'(0) = A((5+sqrt(17))/2)e^0 + B((5-sqrt(17))/2)e^0 = -5
Our 2 equations needed for the unique solution are:
A + B = -1 and A((5+sqrt(17))/2) + B((5-sqrt(17))/2) = -5
if we multiply the first equation by (a2) we obtain:
A((5-sqrt(17))/2) + B((5-sqrt(17))/2) = -1((5-sqrt(17))/2)
Then if we subtract the second equation, B should cancel out yeilding:
(((5-sqrt(17))/2) - ((5+sqrt(17))/2))A = - ((5-sqrt(17))/2) + 5
(1/2)(5 - sqrt(17) -5 -sqrt(17))A = -(1/2)(5-sqrt(17)) + 5
-(sqrt(17))A = -(1/2)(5-sqrt(17)) + 5
2(-(sqrt(17))A) = - (5-sqrt(17)) + 10
2(-(sqrt(17))A) = (sqrt(17) - 5) + 10
2(-(sqrt(17))A) = sqrt(17) + 5
2A = (-sqrt(17) + 5)/(sqrt(17))
A = (-sqrt(17) + 5)/(2*sqrt(17))
And since B = A -1
B = (-sqrt(17) + 5)/(2*sqrt(17)) - 1
y(t) = (-sqrt(17) + 5)/(2*sqrt(17))e^((5 + sqrt(17))t/2) + ((-sqrt(17) + 5)/(2*sqrt(17)) - 1)e^((5 - sqrt(17))t/2)
an approximation would be
y(t) = -1.206e^(4.562t) - (2.106)e^(0.438t)
And as t approches infinity the solution can be seen to decrease.
As it approches negative infinity solution can be seen to increase.
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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:
Using the same method as before:
a = lambda
8a^2 - 6a + 1 = 0
a_1,2 = (-b +,- sqrt(b^2 - 4ac))/(2a)
= (6 +,- sqrt(36 - 4(8)(1)))/(2(8))
a1 = (6 + sqrt(4))/16 = 1/2
or
a2 = (6 - sqrt(4))/16 = 1/4
y(t) = Ae^(t/2) + Be^(t/4)
y '(t) = (A/2)e^(t/2) + (B/4)e^(t/4)
y(1) = Ae^(1/2) + Be^(1/4) = 4
y '(1) = (A/2)e^(1/2) + (B/4)e^(1/4)
A(sqrt(e)) + B(sqrt(sqrt(e))) = 4
minus
(1/2)(A * sqrt(e)) + (1/4)(B * sqrt(sqrt(e))) = 3/2 (Multiply this by 4 before subtraction from the previous)
yeilds:
2(A*sqrt(e)) = 4 - 4(3/2) = -2
A*sqrt(e) = -1
A = -e^(-1/2)
Pluging it back into an equation will let B be solved for:
A(sqrt(e)) + B(sqrt(sqrt(e))) = 4
(-e^(-1/2))(sqrt(e)) + B(sqrt(sqrt(e))) = 4
-1 + Be^(1/4) = 4
Be^(1/4) = 5
B = 5e^(-1/4)
Solution becomes:
y(t) = (-e^(-1/2))e^(t/2) + (5e^(-1/4))e^(t/4)
An approximation:
y(t) = -1.65e^(t/2) + 3.89e^(t/4)
By plugging in a series of numbers we can see how it behaves
3 0.85
2 1.94
1 2.28
-1 2.03
-2 1.75
-3 1.47
From the data we can see as the solution approched infinity it decreases.
As well as it approches negative infinity.
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Very good.
sqrt(sqrt(e)) = e^(1/4), so your equations are
A e^(1/2) + B e^(1/4) = 4
A * 1/2 e^(1/2) + B * 1/4 e^(1/4) = 3/2
Something is wrong in your solution; I believe that when you multiply the second equation by 4 and subtract you end up with - A sqrt(3), not 2 A sqrt(e).
In any case the approximate solution to these equations, which are correct equations, is
A = 1.213
B = 1.558
so the function is
y(t) = 1.213 e^(t/2) + 1.558 e^(t/4).
You can verify that for this function, up to roundoff error,
y(1) = 4
y ' (1) = 3/2.
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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:
I attempted to solve this one using the previous method:
m ( r '' - Omega^2 r) = - k r '
r '' + (k/m) r ' - Omega^2 r) = 0
P(a) = a^2 + (k/m)a + Omega^2 = 0
Quadratic Equation:
a_1,2 = ((-k/m) +,- sqrt((k/m)^2 - 4(1)(omega^2)))/2
(a_1,2)^2 = ((-k/m)^2 +,- (k/m)^2 - 4(1)(omega^2))/4
@&
If you square a quantity of the form
(a + sqrt(b))
you don't be a^2 + b, you get a^2 + 2 a sqrt(b) + b.
Your characteristic equation and its solutions are as you say:
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.
*@
(a1)^2 = (2(k^2/m^2) - 4(omega^2))/4 = ((k^2/m^2) - 2(omega^2))/2
or
(a2)^2 = (-4(omega^2))/4 = -(omega^2)
Not sure if this method worked, or if this type of problem needs a different method entirly.
a1 = sqrt(((k^2/m^2) - 2(omega^2))/2)
a2 = sqrt(-(omega^2))
Making r(t) = Ae^((sqrt(((k^2/m^2)t) - 2(omega^2))/2)) + Be^((sqrt(-(omega^2)))t)
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Good work, though you did have a few errors in details. Still you are using the correct methods to solve the problems and you're in good shape here. Just be as careful as possible with the algebra.
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