Assignment 7

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course Mth 279

6/24 10:00

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*t = e^y*sin(t)

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

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

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

int(e^-y) dy = int(sin(t) - t) dt

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

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

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

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

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

1 = 1 - C

C = 0

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

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1

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Given Solution:

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Self-critique (if necessary):

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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^2*(dy/dt) + 2t = 1

3y^2 * (dy/dt) = 1 - 2t

3y^2 dy = (1-2t) dt

integrate both sides

3y^3/3 = t - 2t^2/2 + C

y^3 = t - t^2 +C

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

-1 = (0-0^2 + C)^(1/3)

C = -1

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

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2

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Given Solution:

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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:

y^3 + sin(y) = 4 - t^2

take derivatives of both sides

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

y’(3y^2 + cos(y)) = -2t

y’ = (-2t)/(3y^2 + cos(y))

when t = 2

y^3 + (2)^2 + sin(y) = 4

y^3 + sin(y) = 0

y^3 = -sin(y)

y = 0

so y(2) = 0

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1

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Given Solution:

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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:

dy/dt = (y^2 + 2y +1)sin(t)

1/(y+1)^2 dy = sin(t) dt

integrate

left side use u sub

letting u = y + 1 therefore du = 1 dy

int(1/u^2)du = -u^-1

so we get

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

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

y = (1/(cos(t) + C)) -1 ; cos(t) + C can’t equal 0

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For values of c between -1 and 1 there would be discontinuities in the y function.

y ' = (y^2 + 2 y + 1) sin(t)

is however continuous for all possible values of y and t. This means, for example, that a direction field could be defined at every point of the y vs. t plane, so that a solution could be found for any initial condition.

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confidence rating #$&*:

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2

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Given Solution:

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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:

1) y’ = -y^2

as y increases the slope of y’ will be a greater negative slope

2) y’ = y^3

as y increases the slope will be positive and will be increasing

3) y’ = y(4-y)

at y = 0 and y=4 the slope would be 0

confidence rating #$&*:

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1

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Given Solution:

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Self-critique (if necessary):

No graphs were provided to match so I tried to explain how the graphs would might look

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&#This looks good. See my notes. Let me know if you have any questions. &#