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course Mth 279
6/20 8:40
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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y(t) = -ln( | cos(t) + t^2/2 |
The absolute value sign is necessary.
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confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
I was unsure about when you take the ln of both side of the equation how to deal with the negative sigh. For example I had ln^(-e^-y) would simplify to be -y or +y(since there would be two - signs???
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The natural log of a negative quantity is undefined.
You would multiply both sides of your equation
-e^-y = -cos(t) - t^2/2 + C
by -1 before taking the natural log.
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Question: 3.2.10. Solve 3 y^2 y ' + 2 t = 1 with initial condition y(0) = -1.
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Your solution:
3y^2*(dy/dt) + 2t = 1
3y^2 * (dy/dt) = 1 - 2t
3y^2 dy = (1-2t) dt
integrate both sides
3y^3/3 = t - 2t^2/2 + C
y^3 = t - t^2 +C
y = (t - t^2 + C)^(1/3)
-1 = (0-0^2 + C)^(1/3)
C = -1
y(t) = (t -t^2 -1)^(1/3)
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Given Solution:
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Self-critique (if necessary):
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Question: 3.2.18. State a problem whose implicit solution is given by y^3 + t^2 + sin(y) = 4, including a specific initial condition at t = 2.
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Your solution:
y^3 + sin(y) = 4 - t^2
take derivatives of both sides
d/dt(y^3 + sin(y) = d/dt(4 - t^2)
y’(3y^2 + cos(y)) = -2t
y’ = (-2t)/(3y^2 + cos(y))
when t = 2
y^3 + (2)^2 + sin(y) = 4
y^3 + sin(y) = 0
y^3 = -sin(y)
y = 0
so y(2) = 0
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
I think this problem you have to kind of work backwards from what we were previously doing.
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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’tequal 0
confidence rating #$&*:
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Given Solution:
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Self-critique (if necessary):
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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
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I believe the graphs are in the text.
In any case you can compare your analysis 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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confidence rating #$&*:
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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
&#Good work. See my notes and let me know if you have questions. &#