Query 05

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course MTH 279

Query 05 Differential Equations*********************************************

Question: 3.2.6. Solve y ' + e^y t = e^y sin(t) with initial condition y(0) = 0.

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Your solution:

dy/dt = e^y*sin(t) - e^y*t

dy/dt*e^(-y) = sin(t) - t

ʃe^(-y)(dy/dt)dt = ʃ(sin(t)-t)dt

-e^(-y) + c_1 = -cos(t) - t^2/2 + c_2

-e^(-y) = -cos(t) - t^2/2 + C

-y = ln(cos(t) + t^2/2 + C)

y = -ln(cos(t) + t^2/2 + C)

y(0) = 0 >> -ln(cos(0) + 0^2/2 + C) = 0

-ln(1 + C) = 0

C = 0

y(t) = -ln(cos(t) + t^2/2)

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: 3.2.10. Solve 3 y^2 y ' + 2 t = 1 with initial condition y(0) = -1.

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Your solution:

3y^2dy/dt = 1-2t

ʃ3y^2dy/dt*dt = ʃ1-2tdt

y^3 + c_1 = t - t^2 + c_2

y^3 = t - t^2 + C

y = (t-t^2+C)^(1/3)

y(0) = -1

-1 = C^(1/3)

-1^3 = C

-1 = C

y(t) = (t-t^2-1)^(1/3)

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: 3.2.18. State a problem whose implicit solution is given by y^3 + t^2 + sin(y) = 4, including a specific initial condition at t = 2.

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Your solution:

d/dt(y^3) + d/dt(t^2) + d/dt(sin(y)) = d/dt(4)

3y^2*y’ + 2t + cos(y)*y’ = 0 ; y(0) = 2

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Good, but your initial condition is at t = 0 rather than t = 2.

*@

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: 3.2.24. Solve the equation y ' = (y^2 + 2 y + 1) sin(t) and determine the t interval over which the solution exists.

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Your solution:

ʃdy/((y+1)^2) = ʃsin(t)dt

-1/(y+1) + c_1 = -cos(t) + c_2

-1/(y+1) = -cos(t) + C

y = 1/(cos(t)-C) - 1

The solution exists at all real value t, such that cos(t)-C ≠ 0.

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: 3.2.28. Match the graphs of the solution curves with the equations y ' = - y^2, y ' = y^3 and y ' = y ( 4 - y).

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Your solution:

There are no graphs listed with this assignment on the website. I found this assignment at http://vhcc2.vhcc.edu/dsmith/GenInfo/qa_query_etc/differential_equations/query_05.htm

I’m not finding any graphs related to this assignment. I can however describe the graphs, if that is acceptable.

ʃdy/y^2 = ʃ-1dt

-1/y + c_1 = -t + c_2

-1/y = -t + C

y = -1/(-t+C) = 1/(t-C)

This produces a graph that tends to 0 as (t+C) increases to ∞ (horizontal asymptote). There is a vertical asymptote as (t+C) approaches 0. It approaches ∞ from the right side, and -∞ from the left side. It also tends to 0 as (t+C) decreases to -∞ (horizontal asymptote).

y’ = y^3

y = (-1/(2t+C))^(1/2)

This solution exists everywhere that (2t+C) ≠ 0 and (2t+C) doesn’t take a positive value. Assuming that C=0 for the moment, the solution would have a vertical asymptote at t = 0, and would cease to exist for any positive t value. As t approaches negative infinity, values tend towards 0.

y ' = y ( 4 - y)

y = 4e^(4(x+C))/(e^(4(x+C))+1)

This solution exists at every value of t. As t approaches infinity, y approaches 4. As t approaches negative infinity, y approaches 0.

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

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