#$&*
course Mth 279
7/15
Query 14 Differential Equations*********************************************
Question: Decide whether y_1 = 3 e^t, y_2 = e^(t + 3) are solutions to the equation y '' - y = 0. If so determine whether the two solutions are linearly independent. If the solutions are linearly independent then find the general solution, as well as a particular solution for which y (-1) = 1 and y ' (-1) = 0.
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Your solution:
1) sub in y_1 and y_2 into equation to see if they = 0
a)y’’_1 - y = (3e^t)’’ - 3e^t = 3e^t - 3e^t = 0
yes y_1 is solution
b) y’’_2 - y = (e^(t+3))’’ - e^(t+3) = e^(t+3) - e^(t+3) = 0
yes y_2 is solution
2) are solution linearly independent? use wronskian
W(t) = |f g; f’ g’|
W(t) = |3e^t e^(t+3); 3e^t e^(t+3)| = (3e^t)(e^(t+3) - (e^(t+3))(3e^t) = 0
since W(t) = 0 , they are NOT linearly independent
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
What happens if W(t) does not equal 0??? That would show linearly independent so what would we do next to find general solution???
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Self-critique rating:
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The general solution would in this case be a linear combination of the linearly independent solutions.
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Question: Decide whether y_1 = e^(-t) and y_2 = 2 e^(1 - t) are solutions to the equation y '' + 2 y ' + y = 0. If so determine whether the two solutions are linearly independent. If the solutions are linearly independent then find the general solution, as well as a particular solution for which y (0) = 1 and y ' (0) = 0.
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Your solution:
1) sub in y_1 and y_2
y_1 = e^-t ; y’_1 = -e^-t ; y’’_1 = e^-t
(e^-t) + 2(-e^-t) + e^-t = 0?
2e^-t - 2e^-t = 0 yes solution
y_2 = 2e^(1-t); y’_2 = -2e^(1-t) ; y’’_2 = 2e^(1-t)
(2e^(1-t)) + 2(-2e^(1-t)) + 2e^(1-t) = 0?
4e^(1-t) - 4e^(1-t) = 0 yes solution
2) linearly independent?
W(t) = | e^-t 2e^(1-t); -e^-t -2e^(1-t)| =
= (-2e^-t*e^(1-t)) - (-2e^-t*e^(1-t)) = 0
Not linearly independent
confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
1
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Given Solution:
&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&
Self-critique (if necessary):
still not confident on what to do it W(t) does not equal 0. It might have been helpful having a question that on the homework that didn’t equal 0.
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Self-critique rating:
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Question: Suppose y_1 and y_2 are solutions to the equation
y '' + alpha y ' + beta y = 0
and that y_1 = e^(2 t). Suppose also that the Wronskian is e^(-t).
What are the values of alpha and beta?
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Your solution:
y_1 = e^(2t) ; y’_1 = 2e^(2t) ; y’’_1 = 4e^(2t)
1) Plug into equation
4e^(2t) + a(2e^(2t)) + b(e^(2t) = 0
2) factor out e^(2t)
e^(2t) (4 + 2a + b) = 0
divide e^(2t) over
4 + 2a + b = 0 (one soln - use wronskian to find y_2)
3) W(t) = |e^(2t) y_2; 2e^(2t) y’_2| = e^(2t)y’_2 - y_2(2e^(2t)) = e^-t
e^(2t)y’_2 - 2e^(2t)y_2 = e^-t
y’ - 2y = e^(-3t)
p(t) = -2 int(p(t)) = -2t
y_2 = Ce^(2t) + e^(2t)int(e^(-2t)*e^(-3t))
how do you integrate? You are left with C constant and no initial condition values???
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The Wronskian is
det ( [ e^(2 t), y_2; 2 e^(2 t), y_2 ' ] = e^(-t)
Thus
y_2 ' e^( 2 t ) - 2 y_2 e^(2 t) = e^(-t).
Multiplying both sides by e^(-2 t) we get
y_2 ' - 2 y_2 = e^(-3 t).
This is a nonhomogeneous linear equation that can be solved for the function y_2.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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#*&!
&#This looks good. See my notes. Let me know if you have any questions. &#