#$&*
course Mth 279
April 30 around 3:50pm.
Query 15 Differential Equations*********************************************
Question: Suppose y1 and y2 are solutions to y '' + 2 t y ' + t^2 y = 0. If y1(3) = 0, y1 ' (3) = 0, y2(3) = 1 and y2 ' (3) = 2, can you say whether {y1, y2} is a fundamental set? If so, is it or isn't it?
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Your solution:
y1(3) = 0
y1 ' (3) = 0
y2(3) = 1
y2 ' (3) = 2
y '' + 2 t y ' + t^2 y = 0
I think I need to use these equations:
y(t0) = c1y1(t0) + c2y2(t0)
y’(t0) = c1y1’(t0) + c2y2’(t0)
OR the Wronskian to identify:
y1(t0)y2’(t0) - y2(t0)y1’(t0) doesn’t = 0
Plug in given sets:
0(2) - 1(0) = 0
W(t) = 0, therefore {y1,y2} is NOT a fundamental set.
Is this right?! I feel like I’m doing something wrong.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@& You can find the Wronskian at t = 3:
W(3) = det [ 0, 1; 0, 2 ] = 0.
The equation is defined for all values of t.
If the Wronskian is zero for one value of t in an interval of definition, then it's zero for all values of t in that interval.
The interval of definition is the entire real axis, so the Wronskian is zero everywhere. So these functions cannot form a fundamental set.*@
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Question: Are y1 = 2 e^(-2 t) cos(t) and y2 = e^(-2 t) sin(t) solutions to the equation
y '' + 4 y ' + 5 y = 0?
What are the initial conditions at t = 0?
Is {y1, y2} a fundamental set?
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Your solution:
y1 = 2 e^(-2 t) cos(t)
y2 = e^(-2 t) sin(t)
Need to find y1’ and y2’:
Y1’ = (-4 cos(t)-2 sin (t)) * e^(-2t)
Y2’ = (cos(t) - 2 sin(t)) * e^(-2t)
y '' + 4 y ' + 5 y = 0 (main equation)
I think I need to use these equations:
y(t0) = c1y1(t0) + c2y2(t0)
y’(t0) = c1y1’(t0) + c2y2’(t0)
OR the Wronskian to identify:
y1(t0)y2’(t0) - y2(t0)y1’(t0) doesn’t = 0 (if it doesn’t equal 0, then {y1,y2} IS a fundamental set)
Plug in given sets:
= [2 e^(-2 t) cos(t)]*[ (cos(t) - 2 sin(t)) * e^(-2t)] - [e^(-2 t) sin(t)]*[ (-4 cos(t)-2 sin (t)) * e^(-2t)]
= [2 cos (t) * (cos (t) - 2 sin (t)) * e^(-4t)] - [-2 sin (t) * (2 cos (t) + sin (t)) * e^(-4t)]
=2e^(-4t)
If t = 0, then
= 2
So, since it’s 2 and not 0, does this mean that it IS a fundamental set?!
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):
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Self-critique rating:
@& Good.*@
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Question: y1_bar = 2 y1 - 2 y2 and y2_bar = y1 - y2. Is {y1_bar, y2_bar} a fundamental set?
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Your solution:
y1bar = 2y1 - 2 y2
y2bar = y1 - y2
Need to find y1bar’ and y2bar’:
y1bar’ = 2
y2bar’ = -1
Use the Wronskian to identify:
y1(t0)y2’(t0) - y2(t0)y1’(t0) doesn’t = 0 (if it doesn’t equal 0, then {y1,y2} IS a fundamental set)
Plug in values:
= (2y1 - 2 y2)*(-1) - (y1 - y2)*(2)
= (-2y1+2y2) - (2y1-2y2)
= (-4y1 + 4y2)
Doesn’t equal zero, therefore {y1_bar, y2_bar} IS a fundamental set
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: Note that y_1_bar = 2 * y_2_bar.
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Self-critique (if necessary):
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Self-critique rating:
@& Good.*@
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Self-critique (if neces