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course Mth 279
2/26 21
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’ = -e^y ( t - sin(t))
divide each side by e^y and integrate
int e^-y y’ = int -t + sin(t)
-e^-y = -t^2 / 2 - cos(t) + c_1
y = -log(t^2 / 2 +cos(t) - c_1
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The value of c_1 is the one that makes the expression satisfy the initial condition. At this point c_1 can be positive or negative, or (as it will turn out) zero.
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y = -log(t^2 / 2 +cos(t) + c_1
y(0) = 0
0 = -log( 0 + 1 + c_1)
0 = -log(1 + c_1)
c_1 = 0
y = -log(t^2 / 2 +cos(t)
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Good.
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Given Solution:
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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 y’ = -2 t + 1
integrate each side with respect to t
y^3 = -t^2 + t + c_1
y = 3rd root [ -t^2 + t + c_1]
y(0) = -1
-1 = 3rd root [ c_1]
c_ 1 = - 1
y = 3rd root [ -t^2 + t - 1]
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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:
??? I do not understand where to begin when (I assume) working backwards through this problem.
y^3 + sin(y) = -t^2 + 4
Do I take the derivative of each side? This would still lead to a y^2 term on the left side.
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You are on the right track. A y^2 term is no problem in a separable equation.
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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:
Divide each side by (y^2 + 2 y + 1)
y’ / (y^2 + 2 y + 1) = sin(t)
integrate each side with respect to t
- 1/ (y + ) = -cos(t) + c_1
y = ( -cos(t) + c_1 + 1) / (cos(t) - c_1)
The solution does not exist when c_1 = cos(t)
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Upon integrating we do get
-1 / (y + 1) = -cos(t) + c.
Solving this for y we get
y = 1 / (cos(t) + c) - 1.
For any value of c between -1 and 1 there would be discontinuities in the y function, since cos(t) takes all values between -1 and 1.
However
y ' = (y^2 + 2 y + 1) sin(t)
is 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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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:
??? There are no graphs in both the copied word document or the online link.
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&#Your work looks good. See my notes. Let me know if you have any questions. &#
course Phy 121
Possible Combinations of VariablesThere are ten possible combinations of three of the the five variables v0, vf, a, Dt and Ds. These ten combinations are summarized in the table below:
1
v0
vf
a
2
v0
vf
dt
3
v0
vf
ds
4
v0
a
dt
5
v0
a
ds
*
6
v0
dt
ds
7
vf
a
dt
8
vf
a
ds
*
9
vf
dt
ds
10
a
dt
ds
If we know the three variables we can easily solve for the other two, using either direct reasoning or the equations of uniformly accelerated motion (the definitions of average velocity and acceleration, and the two equations derived from these by eliminating Dt and then eliminating vf).
Only two of these situations require equations for their solution; the rest can be solved by direct reasoning using the seven quantities v0, vf, a, Dt, Ds, Dv and vAve. These two situations, numbers 5 and 8 on the table, are indicated by the asterisks in the last column.
Direct Reasoning
We learn more physics by reasoning directly than by using equations. In direct reasoning we think about the meaning of each calculation and visualize each calculation.
When reasoning directly using v0, vf, `dv, vAve, `ds, `dt and a we use two known variables at a time to determine the value of an unknown variable, which then becomes known. Each step should be accompanied by visualization of the meaning of the calculation and by thinking of the meaning of the calculation. A 'flow diagram' is helpful here.
Using Equations
When using equations, we need to find the equation that contains the three known variables.
We solve that equation for the remaining, unknown, variable in that equation.
We obtain the value of the unknown variable by plugging in the values of the three known variables and simplifying.
At this point we know the values of four of the five variables.
Then any equation containing the fifth variable can be solved for this variable, and the values of the remaining variables plugged in to obtain the value of this final variable.
Problem
Do the following:
Make up a problem for situation # 4, and solve it using direct reasoning.
Accompany your solution with an explanation of the meaning of each step and with a flow diagram.
Then solve the same problem using the equations of uniformly accelerated motion.
Make up a problem for situation # 5, and solve it using the equations of uniformly accelerated motion.
Problem #4:
v0 a 'dt
A ball is traveling down a incline with a initial velocity of 6 m/s for 12 s at a constant acceleration of 2 m/s^2.
v0 = 6 m/s
'dt = 12 s
a = 2 m/s^2
With what we know we can find the change of 'dv:
'dv = 2 m/s^2 * 12 s = 24 m/s
Now we can find the finial velocity:
vf = 24 m/s + 2 m/s^2(12 s) = 30 m/s
Now we can find the average velocity:
vAve = (6 m/s + 30 m/s) / 2 = 18 m/s
Now we can find the displacement:
'ds = 18 m/s * 12 s = 216 m
Problem #5:
v0 a 'ds
A ball is traveling down a incline with a initial velocity of 4 m/s for a distance covered of 240 m and an acceleration of 16 m/s^2.
v0 = 4 m/s
'ds = 240 m
a = 16 m/s^2
(TF = therefore)
'ds = vAve * 'dt
'dt = 'dv / a
TF 'ds = (vAve * 'dv) / a
'dv = vf - v0
vAve = (vf + v0) / 2
TF 'ds = (vf + v0)(vf - v0) / 2a
'ds = (vf^2 - v0^2) / 2a
2a'ds + v0^2 = vf^2
vf = sqrt(2a'ds + v0^2)
Now enter the known information:
vf = sqrt[2*(16 m/s^2)(240 m) + (4 m/s)^2]
vf = sqrt[7680 m^2/s^2 + 16 m^2/s^2]
vf = sqrt[7696 m^2/s^2]
vf = 87.7 m/s
Now we can find the average velocity:
vAve = (87.7 m/s + 4 m/s) / 2 = 45.9 m/s
Now we can find the clock time:
'dt = (45.9 m/s) / (16 m/s^2) = 2.9 s
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Excellent work.
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