Query 17

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course MTH 279

3/23 730

Query 17 Differential Equations*********************************************

Question: Solve the equation

25 y '' + 20 y ' + 4 y = 0, y(5) = 4 e^-2, y ' (5) = -3/5 e^-2

with y(5) = 4 e^-2 and y ' (5) = -3/5 e^-2.

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Your solution:

y = -e^(-2t/5) + te^(-2t/5)

Using the quadratic formula, I found the repeated root -2/5

So my general solution was therefore y= C_1e^(-2t/5) + C_2te^(-2t/5)

Using the given conditions, I found C_1 = -1 and C_2 = 1

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: Solve the equation

3 y '' + 2 sqrt(3) y ' + y = 0, y(0) = 2 sqrt(3), y ' (0) = 3

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Your solution:

y = 2sqrt(3)e^(-tsqrt(3)/3)) - 3/sqrt(3) te^(-tsqrt(3)/3))

Using the quadratic formula, I found the repeated root -sqrt(3)/3

And using the initial conditions, I found C_1 = 2sqrt(3) and C_2 = -3/sqrt(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: Solve the equation

y '' - 2 cot(t) y ' + (1 + 2 cot^2 t) y = 0,

which has known solution y_1(t) = sin(t)

You will use reduction of order, find intervals of definition and interval(s) where the Wronskian is continuous and nonzero. See your text for a more complete statement of this problem.

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Your solution:

y = C_1sin(t) + C_2ktsin(t) defined on the interval 0

Since y_1 is known, we know that y_2 = usin(t) where u is some function of t.

Taking its first deriv: y' = u'sin(t) + ucos(t)

second deriv: y'' = u''sin(t) + u'cos(t) - usin(t)v+ u'cos(t)

Substituting into the original DE and simplifying with a few basic trig ID's, we get:

u''sin(t) = 0

which gives us u = kt

so our linear combination is : y = C_1sin(t) + C_2ktsin(t)

Computing the Wronskian, we find it to be = csin^2(t) which is continuous everywhere and is zero at t = 0, pi, 2 pi.... Which tells us the largest interval that the solution exists on is an interval such as 0

????? I didn't know if there was a better or more general way to write the interval.

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You could say that the solution applies on any interval of the form

(n * pi, (n+1) * pi).

*@

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"

Self-critique (if necessary):

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Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

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&#Good responses. See my notes and let me know if you have questions. &#