#$&*
course Mth 279
7/16
Query 17 Differential Equations*********************************************
Question: Solve the equation
25 y '' + 20 y ' + 4 y = 0, y(5) = 4 e^-2, y ' (5) = -3/5 e^-2
with y(5) = 4 e^-2 and y ' (5) = -3/5 e^-2.
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Your solution:
y=Ae^(rt)
y=rAe^(rt)
y=r^2Ae^(rt)
25 y '' + 20 y ' + 4 y = 0
25r^2 + 20r + 4 = 0
(5r+2)(5r+2) = 0
r = -2/5
y_1 = Ae^(-2/5 t)
trick y_2 = Bte^(-2/5 t)
gen soln:
y(t) = Ae^(-2/5 t) + Bte^(-2/5 t)
y(t) = -2/5Ae^(-2/5 t) + Be^(-2/5 t) - 2/5Bte^(-2/5 t)
y(5) = Ae^(-2/5 * 5) + B(5)e^(-2/5*5)
4e^-2 = Ae^-2 + 5B^-2
4 = A + 5B
y(5) = -2/5Ae^(-2/5 *5) + Be^(-2/5 *5) - 2/5B(5)e^(-2/5 *5)
-3/5e^(-2) = -2/5Ae^-2 + Be^-2 -2Be^-2
-3/5 = -2/5A - B
solve systems of equations
multiply second equation by 5 and add the two equations
1 = -1A
A = -1
so B will be 4 = -1 + 5B
B =1
y(t) = -e^(-2/5 * t) + te^(-2/5 * t)
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Question: Solve the equation
3 y '' + 2 sqrt(3) y ' + y = 0, y(0) = 2 sqrt(3), y ' (0) = 3
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Your solution:
3r^2 + 2sqrt(3)r + 1 = 0
solving for r using quadratic formula
r = -2sqrt(3) +- sqrt((2sqrt(3))^2 - 4(3))/2(3)
r = -2sqrt(3) +- sqrt(12-12)/6 = -2sqrt(3)/6 = -0.577
using trick to get y_2
y_2 = Bte^-0.577t
y(t) = Ae^(-0.577t) + Bte^(-0.577t)
y(0) = Ae^0 + B(0)e^0 = 2sqrt(3)
A = 2sqrt(3)
y = -0.577Ae^(-.577t) + Be^(-.577t) -.577Bte^(-.577t)
y(0) = -.577A + B 0 .577B(0) = 3
3 = -.577(2sqrt(3)) + B
3 = -2 + B
B = 5
y(t) = 2sqrt(3)e^(-.577t) + 5te^(-.577t)
@&
Good, but you introduce errors when you approximate sqrt(3) / 3 as .577. When in doubt, use exact values. Your solution would be more correct expressed as
y(t) = 2 sqrt(3) e^(-sqrt(3) / 3 * t) + 5 t e^(-sqrt(3) / 3 * t).
*@
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Question: Solve the equation
y '' - 2 cot(t) y ' + (1 + 2 cot^2 t) y = 0,
which has known solution y_1(t) = sin(t)
You will use reduction of order, find intervals of definition and interval(s) where the Wronskian is continuous and nonzero. See your text for a more complete statement of this problem.
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Your solution:
reduction of order
assume y_2 = x_1*u(t)
y_2 = sin(t)u(t)
y_2 = sin(t)u + ucos(t)
y_2 = sin(t)u + 2cos(t) u -sin(t)u
sub into equation
0 = (sin(t)u + 2cos(t) u -sin(t)u) - 2cot(t)(sin(t)u + ucos(t)) + (1+2cot^2(t))(sin(t)u)
simplifying
0 = sin(t)u
Let v = u and v = u
0 = sin(t)v
v = 0
???if v = 0 and v=u then u would = 0???
I dont think thats right. Not sure what to do next???
@&
Good. The rest falls into place as follows:
y_2 ' = u ' * y_1 + u * y_1 ' = u ' sin(t) + u cos(t)
and
y_2 '' = u'' * y_1 + 2 u ' * y_1 ' + u y_1 ''
= u '' * sin(t) + 2 u ' * cos(t) - u * sin(t).
It's a little easier to keep track of things by writing the equation in terms of u, y_1 and their derivatives. Doing so your equation
0 = (sin(t)u + 2cos(t) u -sin(t)u) - 2cot(t)(sin(t)u + ucos(t)) + (1+2cot^2(t))(sin(t)u)
can be written
u'' * y_1 + 2 u ' * y_1 ' + u y_1 '' - 2 cot(t) (u ' * y_1 + u * y_1 ') + (1 + 2 cot^2(t)) * u y_1 = 0
which can be rearranged to yield
[ u y_1 '' - 2 cot(t) * u y_1 ' + (1 + 2 cot^2(t)) * u y_1 ] + u '' y_1 + u ' ( 2 y_1 - 2 cot(t) y_1) = 0.
The terms in brackets can be expressed as u ( y_1 '' - 2 cot(t) * y_1 ' +(1 + 2 cot(t)^2) y_1); since y_1 is a solution to our original equation these terms therefore add up to zero.
This leaves us with
u '' + 2 u ' y_1 (1 - cot(t)) = 0.
We can now substitute sin(t) for y_1 to get
u '' + 2 u ' sin(t) ( 1 - cot(t)) = 0.
This is a first-order linear equation in u '. Letting v = u ' to make this clear you get an equation you can solve for v.
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#$&*
07-17-2014
&#Good responses. See my notes and let me know if you have questions. &#