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course MTH 279
3/1 9 pm
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:
Yes, both are solutions (taking the 1st and 2nd derivs and plugging them in yielded 0). Yes, solutions are linearly independent (Wronskian = 6e^t -2e^(t+3)).
General soluton: y"" = 3C_1 e^t + C_2 e^(t+3)
Particular solution using initial conditions: e/2 * e^t - 1/(e^2) * e^(t+3)
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
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Question: Decide whether y_1 = e^(-t) and y_2 = 2 e^(1 - t) are solutions to the equation y '' + 2 y ' + y = 0. If so determine whether the two solutions are linearly independent. If the solutions are linearly independent then find the general solution, as well as a particular solution for which y (0) = 1 and y ' (0) = 0.
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Your solution:
Yes, both are solutions.
No, solutions aren't linearly independent. Wronskian = -2e^(1-2t) + 2e^(1-2t) = 0
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Right. Note that 2 e^(1 - t) = 2 e * e^(-t), making the second function a scalar multiple of the first. You can think about why this makes the Wronskian zero (if you find the Wronskian of functions y_1 and y_2 = c * y_1, this becomes clear).
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
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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:
alpha = 1
beta = -6
To find alpha and beta, I first found the ""general"" Wronskian and its derivative.
W = e^-t = y_1y_2' - y_2y_1'
W' = -e^-t = y_1y_2'' - y_2y_1''
I then used given information and known relationships to sub out y_1'' and y_2''. Namely, y_1 '' = -alpha y_1 ' - beta y_1 and y_2 '' = -alpha y_2 ' - beta y_2
Simplifying the new equation, I found W' = -alphaW
Which lead me to find alpha = 1 with the given information. I could then plug into y_1 '' + alpha y_1 ' + beta y_1 = 0 to find beta.
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Very good solution, more sophisticated than the following, which I've included for comparison:
Let y_2 be the unknown solution.
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 which, if necessary, can be solved using integrating factor e^(- 2 t). However by inspection it is not difficult to see that a solution is
y_2 = e^(-3 t) / 5.
Thus our general solution is
y = A e^(2 t) + B e^(-3 t).
It follows that our characteristic equation has solutions r = 2 and r = -3. The characteristic equation is therefore
(r - 2) ( r + 3) = 0; expanding we get
r^2 + r - 6 = 0.
Our original equation was therefore
y '' + y ' - 6 = 0,
so that
alpha = 1 and beta = -6.
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confidence rating #$&*:
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&#This looks good. See my notes. Let me know if you have any questions. &#