Query 17

#$&*

course MTH 279

4/6 3:41 pm I submitted to other queries that on Friday that haven't been posted

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.

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Your solution: 

 r^2+ 4/5*r + 4/25= 0

real repeated root

r_1= -0.4

y_1= e^(-0.4t)

By equation 4.5-6

y= c1e^(-0.4t) +c2*t*e^(-0.4t)

Sub in to solve for c1&c2

c1+c2=4

-2/5 c1- c2 = -3/5

c1 = 5.667

c2 = -1.667

Part solution:

y = 5.667e^(-0.4t) - 1.667e^(-0.4t)

confidence rating #$&*:

@& You might have overlooked the t term in y and y ', when plugging in t = 5.

See if you agree with the following:

Our characteristic equation is

25 r^2 + 20 r + 4 = 0

with repeated solution r = -2/5

yielding solution set

{e^(-2/5 t), t e^(-2/5 t) }

and general solution

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

satisfying

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

These conditions lead to the equations

A e^-2 + 5 B e^-2 = 4 e^-2

-2/5 A e^(-2) - 2/5 * 5 B e^-2 + B e^-2 = -3/5 e^-2,

or dividing both equations through by e^-2, and multiplying through the second by 5

A + 5 B = 4

-2 A - 5 B = -3.

The solution is A = 1, B = 3/5, giving us the solution

y(t) = e^(-2/5 t) - 3/5 t e^(-2/5 t).

*@

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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:  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: 

 r^2 + 2/3* sqrt 3*r + 1/3= 0

real repeated:

r_1=- sqrt 3/3

 y = c1e^(- sqrt 3/3t) + c2te^(- sqrt 3/3t)

sub and solve for c1 and c2

c1+c2= 2 sqrt 3

- sqrt 3/3 c1 + c2 = 3

c1 = 0.294

c2= 3.17

y = 0.294e^(- sqrt 3/3t) + 3.17t*e^(- sqrt 3/3t)

@& When you plug in t = 0 the second term of y(t) is zero, because of the factor t. So c_1 = 2 sqrt(3).

See if you agree:

The characteristic equation is

3 r^2 + 2 sqrt(3) r + 1 = 0

with repeated solution

r = (-2 sqrt(3) +- sqrt( 12 - 12 ) ) / 6 = -sqrt(3)/3.

This yields fundamental set

{e^(-sqrt(3) / 3 * t), t e^(-sqrt(3) / 3 * t) }

with general solution

y(t) = A e^(-sqrt(3) / 3 * t) + B t e^(-sqrt(3) / 3 * t).

y(0) = A = 2 sqrt(3)

y ' (0) = -sqrt(3) / 3 * A + B = 3.

Substituting A = 2 sqrt(3) into the second equation we get

-6 + B = 3

so that

B = 9.

Our solution is therefore

y(t) = 2 sqrt(3) e^(-sqrt(3) / 3 * t) + 9 t e^(-sqrt(3) / 3 * t).

*@

 

confidence rating #$&*:

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Given Solution: 

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Self-critique (if necessary):

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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: 

y2= u(t)* sin t

y2' = u'*sin t + u*cos t

y2''= u''* sin t + u'*cos t + u'*cos t - u sin t

sub in

(u''* sin t + u'*cos t + u'*cos t - u sin t) - 2*(cos t/sin t)(u'*sin t + u*cos t)+(1+2(cos t/sin t)^2)(u*sin t) = 0

Everything but the u'' reduces to 0

u'' sin t = 0

u''= 0

Don't really need reduction of order to do this but sub

v = u'

v' = u''

v' sin t = 0

v'= c1= u'

u' = c1t + c2

y2 = (c1*t+c2)*sin t

I'm going to assume that c1 = 1 and c2 = 0 based on y1

y1'= cos t

y2'= sin t + t cos t

W = sin t( sin t + t cos t) - sin t * cos t = sin^2 t

W is continuous everywhere but is nonzero when only sin t is nonzero which is when t = n*π, where n = +Z

Intervals of solutions occur between n and n±1, i.e. -π < t < π, π < t < 2π, etc

The original eq is written in the desired form and

p(t) = -2 cot t, which exists everywhere except where sin t = 0 and that matches the intervals just given

q(t) = 1+2 cot^2 t, which also exists on the same intervals

 

confidence rating #$&*:8232;

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Given Solution: 

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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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#$&*

04-05-2011

@& Good, but when evaluating your constants you might be ignoring the t factor in the second function in the repeated-roots solution. See my notes to be sure.*@