#$&*
course MTH 279
2/5 7 pm
Query 05 Differential Equations*********************************************
Question: 3.2.6. Solve y ' + e^y t = e^y sin(t) with initial condition y(0) = 0.
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Your solution:
y = -ln( cos(t) + t^2/2 )
Separate your variables by dividing eveything by e^y and treating dy/dt like a fraction:
e^(-y) dy = sin(t) - t dt
Integrate both sides:
-e^(-y) = -cos(t) - .5t^2 + C
Solve for y:
y = -ln( cos(t) + t^2/2 + C)
Satisfy the initial condition to find C:
C = 0
confidence rating #$&*:
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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: 3.2.10. Solve 3 y^2 y ' + 2 t = 1 with initial condition y(0) = -1.
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Your solution:
y = cubert (t - t^2 -1)
Separate your equation:
3y^2 dy = 1 - 2t dt
Integrate each side:
y^3 = t - t^2 + C
Solve for y and satisfy the initial condition (not neccessarily in that order. I think sometimes it's easier to plug in to some of the implicitly defined functions).
C = -1
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary): OK
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Self-critique rating: OK
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Question: 3.2.18. State a problem whose implicit solution is given by y^3 + t^2 + sin(y) = 4, including a specific initial condition at t = 2.
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Your solution:
(3y^2 + cos(y) ) y' +2t = 0 and initial condition y(2) = 0
Take the derivative of the original equation:
3y^2 + 2t + cos(y) = 0
Put a y' in there.
@&
You do need that, because
3y^2 + 2t + cos(y) = 0
while on the right track is not a differential equation.
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To find initial condition, you can either solve the equation again, keeping in mind you want C = 4, or find the y value that would make the corresponding t value of the original equation equal to 4.
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
????? I wasn't sure if there was a specific place you needed to put y' in the equation or not.
@&
The equation would be
(3 y^2 + cos(y)) dy + 2 t dt, so that
dy/dt = -2 t / (3 y^2 + cos(y) )
The equation could therefore be written as
y ' = -2 t / (3 y^2 + cos(y) )
or as
(3 y^2 + cos(y) ) y ' = -2 t
or as
(3 y^2 + cos(y) ) y ' + 2 t = 0
or in a variety of other ways.
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Self-critique rating: 3
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Question: 3.2.24. Solve the equation y ' = (y^2 + 2 y + 1) sin(t) fand determine the t interval over which he solution exists.
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Your solution:
y = [-1/(-cos(t) + C)] - 1
@&
Good.
Since C is a constant, which could take any value positive or negative, you would simplify this just a bit to get
y = (1 /(cos(t) + C) ) - 1.
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The t interval includes all reals except (npi)/2 when C = 0, because any (npi)/2 value when C = 0 would make the denominator 0.
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In fact, for and C with | C | <= 1 there would be excluded values.
If | C | > 1 this would no longer be an issue.
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confidence rating #$&*:
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution:
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Self-critique (if necessary):
I'm confindent in my answer, but not the way I expressed the interval.
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Self-critique rating: 3
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Question: 3.2.28. Match the graphs of the solution curves with the equations y ' = - y^2, y ' = y^3 an dy ' = y ( 4 - y).
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Your solution:
y ' = - y^2 graph C
y ' = y^3 graph A
y ' = y ( 4 - y) graph B
confidence rating #$&*:
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Given Solution:
@&
Compare your thinking with the following:
For each function you need to analyze the behavior of the right-hand side as you move around in the y vs. t plane.
-y^2 is always negative unless y = 0, in which case -y^2 is zero, and the magnitude of y^2 increases as an increasing rate as you move away from y = 0.
y^3 is positive for positive y, negative for negative y, and zero for y = 0. The magnitude of y^2 increases as an increasing rate as you move away from y = 0.
y ( 4 - y_) is positive for 0 < y < 4, zero when y = 0 or y = 4, and negative everywhere else.
This results in the following slope-field behaviors:
y ' = - y^2 will result in a direction field that gets steeper as you move away from the t axis, since y^2 increases in magnitude at an increasing rate as you move away from t = 0.
On the t axis y is zero so the direction field is horizontal. For y not equal to zero, since y^2 is positive, - y^2 is always negative. So the field lines all have negative slopes, and get steeper as you move away from the t axis.
y ' = y^3 will again result in a direction field that gets steeper as you move away from the t axis, since y^3 increases in magnitude at an increasing rate as you move away from t = 0.
On the t axis y is zero so the direction field is horizontal.
Above the t axis y is positive so y^3 is positive, so the field lines all have positive slopes, and get steeper as you move away from the t axis..
Below the t axis y is negative so y^3 is negative, so the field lines all have negative slopes, and get steeper as you move away from the t axis..
y ' = y ( 4 - y) is positive between the t axis and the line y = 4, negative above that line. The closer you get to y = 4 the less steep the slopes.
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&#Your work looks good. See my notes. Let me know if you have any questions. &#