#$&*
course Mth 279
7/22 9:21 pm
Query 17 Differential Equations*********************************************
Question: Solve the equation
25 y '' + 20 y ' + 4 y = 0, y(5) = 4 e^-2, y ' (5) = -3/5 e^-2
with y(5) = 4 e^-2 and y ' (5) = -3/5 e^-2.
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Your solution:
25y^2 + 20y +4 = 0
(5y +2)^2 = 0
y = c1 e^(-2/5 t) + c2 t e^(-2/5 t)
NOTE, because the equation is continuous at all points, I was able to simply multiply t by y1 to get y2
If it was not continuous, I would have used reduction of order
initial conditions:
4e^(-2) = c1 e^(-2) + 5c2 e^(-2)
@&
y = c1 e^(-2/5 t) + c2 t e^(-2/5 t)
so
y(5) = y = c1 e^(-2t) + c2 * 5 * e^(-2).
*@
4 = c1 + 5c2
c1 = 4-5c2
y' = -2/5 c1 e^(-2/5 t) + c2 e^(-2/5 t) - 2/5 t e^(-2/5 t)
-3/5 e^(-2) = -2/5 c1 e^(-2) + c2 e^(-2) - 2e^(-2) c2
-3/5 = -2/5 c1 - c2
-3/5 = -2/5 (4-5 c2) - c2
-3/5 = -8/5 + 10/5 c2 - c2
1 = 2c2 - c2
c2 = 1
plugging into equation for c1
4-5 = c1
c1 = -1
y = -e^(-2/5 t) + t e^(-2/5 t)
@&
Good, but your constants aren't quite right; check my note.
*@
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Given Solution:
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Question: Solve the equation
3 y '' + 2 sqrt(3) y ' + y = 0, y(0) = 2 sqrt(3), y ' (0) = 3
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Your solution:
3y^2 + 2sqrt(3) y + 1 = 0
(sqrt(3) y +1)^2 = 0
because this equation exists at all points, I don't have to go through reduction
y = c1 e^(-1/sqrt(3) t) + c2 t e^(-1/sqrt(3) t)
initial conditions
2 sqrt(3) = c1 + 0
c1 = 2*sqrt(3)
y' = -1/sqrt(3) c1 e^(-1/sqrt(3) t) + c2 e^(-1/sqrt(3) t) - 1/sqrt(3) c2 t e^(-1/sqrt(3) t )
initial conditions
3 = -1/sqrt(3) c1 + c2
3 = -1/sqrt(3) (2* sqrt(3)) + c2
3 = -2 + c2
c2 = 5
y = 2*sqrt(3) e^(-1/sqrt(3) t) + 5t e^(-1/sqrt(3) t)
@&
You missed that factor t again. Otherwise very good.
*@
confidence rating #$&*:
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Given Solution:
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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:
because this cannot easily be factored and does not exist everywhere, I must use method of reduction
y2 = y1*u
y2 = u sin(t)
y2' = u' sin(t) + ucos(t)
y2'' = u'' sin(t) - u sin(t)
@&
y2'' = ( u' sin(t) + ucos(t) ) ' = u '' sin(t) + u ' cos(t) + u ' cos(t) - u sin(t) = u '' + 2 u ' cos(t) - u sin(t)
*@
plugging into original equation
u'' sin - u sin - 2 cos/sin (u' sin + u cos) + (1 + 2cos^2/sin^2 ) usin = 0
u'' sin - usin - 2u'cos - 2 u cos^2/sin + usin + 2u cos^2/sin = 0
u'' sin - 2u cos = 0
using v = u' to make it easier to solve
v' sin - 2v cos = 0
separable first order
v' 1/(2v) = cos/sin
1/2 lnv = ln(sin) + C1
1/2 v = sin(t) + C1
v = 2 sin(t) + C1
*** I don't know what happens with these constants. The book gives an explanation but I don't understand how the constants are eliminated
u = -2 cos(t) + C2
to make sure the solution is right, I do the wronskian
W = [ sin(t) -2cos(t),
cos(t) 2sin(t)]
W = 2sin^2(t) + 2cos^2(t)
W = 2 which does not equal 0!
solution becomes
y = c1 sin(t) + c2 cos(t)
The first equation exists everywhere except integer multiples of pi (pi, 2pi, 3pi, etc.)
The solution exists everywhere, and the wronskian exists everywhere
@&
Compare with the following:
We begin by letting y_2(t) = u(t) * y_1(t), so that
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).
Our equation becomes
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.
or substituting y_1 = sin(t)
u '' + 2 u ' sin(t) ( 1 - cot(t)) = 0.
cot(t) = cos(t) / sin(t), and our equation for u becomes
u '' + 2 (sin(t) - cos(t) ) u' = 0.
Letting v = u ' our equation is
v ' + 2 ( sin(t) - cos(t) ) v = 0.
This is a first-order linear equation, solvable for v. The solution is integrated to find u, and our solution is y = u * y_1 = u * sin(t).
*@
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
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#$&*
&#This looks good. See my notes. Let me know if you have any questions. &#