#$&*
Mth 279
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Assignment 16 Question
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Here is a copy of my response to the homework question and your response to my initial question:
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?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
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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What you explained to do is exactly what I did initially. I had my equation y'_2 - 2y_2 = e^(-3t)
When you solve for y_2 using the nonhomogeneous format, you end up getting a constant C and you aren't giving initial conditions to solve for C.
Solving using nonhomogeneous your general solution would be y_2 = Ce^(2t) + e^(2t) * int(e^(-2t)*e^(-3t))
taking integral you get int(e^(-5t)) = -1/5*e^(-5t)
y_2 = Ce^(2t) + (e^(2t))(-1/5)e^(-5t)
y_2 = Ce^(2t) -1/5*e^(-3t)
so how do you solve for C???
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The bottom line is that you don't have to solve for C, since e^(2 t) is already a solution to the original equation.
The nonhomogeneous equation has solution 1/5 e^(-3t). There is no integration constant infolved.
The details:
The equation is
y_2 ' - 2 y_2 = e^(-3t)
Multiplying both sides by the integrating factor e^(-2t) we get
e^(-2t) * y_2 ' - 2 e^(-2t) y_2 = e^(-3t) * e^(-2t)
which we express as
(y_2 e^(-2t)) ' = e^(-5 t).
Integrating both sides we get
y_2 e^(-2t) = -1/5 e^(-3 t) + C
so that
y_2 = -1/5 e^(-3 t) + C e^(2 t).
Note that e^(2 t) is already a solution of the equation, so the C e^(2 t) adds nothing to the general solution and may be discarded.
We conclude that the general solution to the original equation is
y(t) = A e^(2 t) + B e^(-3 t).
From this we can find the values of alpha and beta.
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Also if what I did was wrong please help to explain what I need to do. I know once I find whatever y_2 is then I am able to take 1st and 2nd derivatives to fin alpha and beta. I just cant figure out y_2???
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Good question. Also, commendable persistence on this problem.
Check my note.
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#$&*
Mth 279
Your 'question form' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Question Form_labelMessages **
Assignment 17 Question
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Below is a copy of the homework question, my response and your response:
Question: Is {e^t, 2 e^(-t), sinh (t) } a fundamental set on the interval (-infinity, infinity)?
YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY
Your solution:
sinh(t) = (e^t - e^-t)/2
W(t) = |y_1, y_2, y_3; y’1, y’2, y’3; y’’1, y’’2, y’’3|
Not sure what W(t) equals in a 3x3 matrix in general form
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You just keep adding higher-order derivatives:
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 ''] ).
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I am not sure what the determinate is equation to???
Will there be something like this question on the test or will you stick with something along the lines of |y_1, y_2; y'_1, y'_2|???
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You can certainly expect that you may have to do determinants of higher-order equations on a test.
If your question is how to find this determinant, you should review the topic. There are numerous resources on the Web.
The determinant of this particular matrix is
+ y_1 det [y_2 ', y_3 ' ; y_2 '', y_3 '']
- y_2 det[ y_1 ' , y_3 ' ; y_1 '', y_3 '' ]
+ y_3 det [y_1 ', y_2 '; y_1 '', y_2 '' ]
The lower-order determinants are just 2 x 2 and I asusme you know how to find them.
Note also that you will use determinants heavily in the upcoming chapter on systems of differential equations. So if you aren't familiar with higher-order determinants you will definitely need a review of them, as well as how they arise in the solution of systems of linear equations.
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