#$&*
course Mth 279
4/10 11
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:
No it’s not fundamental set because evaluating the Wronskian at t = t_0 we get
y_1(t_0)*y’_2(t_0) - y’_1(t_0)*y_2(t_0)
=0*2 - 0*1 = 0
Wronskian must be nonzero to show {y_1, y_2} forms 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:
@& The complete solution must indicate the domain of the solution. Here's a complete solution, identical to yours except for the addition of the last paragraph.
The Wronskian is
det ( [y_1, y_1 ' ; y_2, y_2 ' ] ).
At t = 3 the Wronskian is therefore
det( [ 0, 0; 1, 2 ] ) = 0.
There is no restriction on the domain of the solution function, so the Wronskian is always zero. It follows that the set {y_1, y_2} is not 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:
For y1 = 2 e^(-2 t) cos(t) and y2 = e^(-2 t) sin(t) and the equation y '' + 4 y ' + 5 y = 0, we first test to see if 2 e^(-2 t) cos(t) is solution to equation…
y '' + 4 y ' + 5 y = 0, plugging in our y1 value we get
[8e^(-2t)sin(t) + 6e^(-2t)cos(t)] + 4*[-4e^(-2t)cos(t) - 2e^(-2t)sin(t)]
+ 5[2 e^(-2 t) cos(t)] = 0, we get
8e^(-2t)sin(t) + 6e^(-2t)cos(t) - 16e^(-2t)cos(t) - 8e^(-2t)sin(t) + 10e^(-2t)cos(t)
= 0, so yes this is solution
For y2 = e^(-2 t) sin(t) we get,
y '' + 4 y ' + 5 y = 0, plugging in our y2 value we get
[3e^(-2t)sin(t) - 4e^(-2t)cos(t)] + 4*[e^(-2t)cos(t) - 2e^(-2t)sin(t)]
+ 5[e^(-2 t) sin(t)] = 0, we get
3e^(-2t)sin(t) - 4e^(-2t)cos(t) + 4e^(-2t)cos(t) - 8e^(-2t)sin(t) + 5e^(-2t)sin(t)
= 0, so yes this is solution
Init cond at t = 0 are
y1 = 2e^-2*.54 = approx. e^-2, y2=e^-2*.8415=approx. e^-2(Both are approximations)
Calculating Wronskian at t=0 we get nonzero value so yes y1 and y2 are 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:
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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:
To see if {y1_bar, y2_bar} a fundamental set we construct the corresponding (2X2) matrix knowing that y1_bar = a_11*y1 + a_21*y2 and y2_bar = a_12*y1 - a_22y2 so that
A=| 2 1|
| -2 -1|(supposed to represent a matrix even though it does not look like it)
Determinant of matrix is
(2*-1) - (-2*1) = -2 - (-2) = -2+2 = 0, because derterminant is zero we know {y1_bar, y2_bar} is not a fundamental set.
@& Good. Compare with the following, in which the Wronskian is worked out in detail:
The Wronskian is
[ y_1_bar, y_2_bar; y_1_bar ' , y_2_bar ' ]
= [ (2 y_1 - 2 y_2), (y_1 - y_2); (2 y_1 - 2 y_2) ' , (y_1 - y_2) ' ]
= [ (2 y_1 - 2 y_2), (y_1 - y_2) ; (2 y_1 ' - 2 y_2 ' ), (y_1 ' - y_2 ') ]
= (2 y_1 - 2 y_2) * (y_1 ' - y_2 ' ) - (2 y_1 ' - 2 y_2 ' ) * (y_1 - y_2 )
= 2 y_1 y_1 ' - 2 y_1 y_2 ' - 2 y_2 y_1 ' + 2 y_2 y_2 ' - ( 2 y_1 ' y_1 - 2 y_1 ' y_2 - 2 y_2 ' y_1 + 2 y_2 ' y_2 )
= 2 y_1 y_1 ' - 2 y_1 y_2 ' - 2 y_2 y_1 ' + 2 y_2 y_2 ' - 2 y_1 ' y_1 + 2 y_1 ' y_2 + 2 y_2 ' y_1 - 2 y_2 ' y_2
You can check to see that each of the first four terms is matched by its negative among the last four terms, giving result zero.
*@
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):
Because one is mult of other tells us that they linearly dependent of each other right??
I believe I still have the right answer.
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Question: Is {e^t, 2 e^(-t), sinh (t) } a fundamental set on the interval (-infinity, infinity)?
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Your solution:
sinh(t) = (1/2)(e^t - e^-t), so we see that sinh(t) could be produced by a combination of e^t and e^-t. I kind of confused but I believe that this is a solution, would we not need more info like the init value problem?? I’m pretty sure all are defined from -infinity to +infinity and if so and not knowing y’’ + p(t)y’ + q(t)y, then I’m going to say that this would be a fundmental set on interval (-infinity, infinity). Unless we are to take the Wronskian of the three variables???
Confidence rating:
@&
The Wronskian of the 3-element set {y_1, y_2, y_3} is det ( [ y_1, y_2, y_3; y_1 ' , y_2 ', y_3 '; y_1 '', y_2 '', y_3 ''] ).
Using sinh(t) = (e^t - e^(-t) ) / 2 (and perhaps noting that this function is a linear combination of the first two, which does not bode well for a nonzero Wronskian) we have
det [ e^t, e^(-t), (e^t - e^(-t)) / 2; e^t, -e^(-t), (e^t + e^(-t)) / 2; e^t, e^(-t), (e^t - e^(-t)) / 2 ].
If we carefully expand this determinant you will find that all the terms pair up and cancel, giving the expected result that the determinant is zero.
On t > 0, | t | = t so the two functions y = t and y = | t | are identical, hence not linearly independent.
On -infinity < t < infinity, however, the equation
c_1 t + c_2 | t | = 0
would imply that for t > 0
c_1 t + c_2 t = 0,
and hence that c_2 = - c_1;
however for t < 0 our equation would imply that
c_1 t + c_2 ( -t ) = 0,
hence that c_2 = c_1.
Unless c_1 = c_2 = 0, this is a contradiction and we conclude that the two functions are linearly independent.
*@
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Given Solution:
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Self-critique (if necessary):
If we were to take cross product of the three variables and there derivatives(I’m not sure how you would set that up) which the cross product would produce determinant, would that be the approach to soAlving this equation???
@& It's simpler than that.
The cross product is defined only in 3 dimensions, hence would not be a good model for a computation that needs to be defined in any number of dimensions.
In any case this was a very good question.*@
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#*&!
&#Good responses. See my notes and let me know if you have questions. &#