Query 13

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course Mth 279

7/15/13 8:45pm

Query 13 Differential Equations*********************************************

Question:  Find the largest interval on which the equation

y '' + y ' + 3 t y = tan(t)

has a solution, with the initial conditions y(pi) = 1 and y ' (pi) = -1.

 

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Your solution: 

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 y '' + y ' + 3 t y = tan(t)

tan(t) is undefined at -pi/2, pi/2, 3*pi/2, etc

largest solution that includes pi is (pi/2, 3*pi/2)

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 Confidence rating:  

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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:  Find the largest interval on which the equation

t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0

has a solution, with the initial conditions y(1) = 0, y ' (1) = 1.

 Your solution: 

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t y '' + sin(2 t) / (t^2 - 9) y ' + 2 y = 0

y"" + sin(2t)/(t(t^2 - 9) y' + 2y/t = 0

t cannot = 0, -3, 3

Largest interval that includes 1 is (-3, 3)

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confidence rating #$&*:232; 

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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:  Tell whether each of the following is increasing or decreasing, and whether concave down or concave up, in the vicinity of the initial point:

• y '' + y = 2 - sin(t), y(0) = 1, y ' (0) = -1.

• y '' + y = - 2 t, y(0) = 1, y ' (0) = -1.

• y '' - y = t^2, y(0) = 1, y ' (0) = 1.

• y '' - y = - 2 cos(t), y(0) = 1, y ' (0) = 1.

 

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Your solution: 

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y '' + y = 2 - sin(t)

 Decreasing because first derivative y'(0) is negative

Concave up because y"" is positive

y '' + y = - 2 t

Decreasing because first derivative y'(0) is negative

Concave down because y"" is negative when you plug in y(0) = 1

y '' - y = t^2

Decreasing because first derivative y'(0) is negative

Concave up because y"" is positive

y '' - y = - 2 cos(t)

 Increasing because first derivative y'(0) is positive

 Concave down because y"" is negative when you plug in y(0) = 1

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confidence rating #$&*:

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