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course Mth 279
2/16 17
Section 2.2*********************************************
Question: 1. Solve the following equations with the given initial conditions:
1. y ' - 2 y = 0, y(1) - 3
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Your solution:
y’ - 2 y = 0 , y(1) = 3
y’ = 2 y
y’ / y = 2 divide each side by y to get all y’s on one side
int (y’ / y ) = int ( 2 ) take integral of each side
ln y = 2 x + c_1
y = e^(2 x + c_1)
y = c_1 * e^(2 x) rearrange
Use the initial condition y(1) = 3 and plug it into above equation
3 = c_1 * e^(2 * 1)
3 = c_1 * e^2
c_1 = 3 / e^2
Plug this c_1 value into y = c_1 * e^(2 x)
y = (3 / e^2) * e ^ 2 x
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Given Solution:
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Question: 2. t^2 y ' - 9 y = 0, y(1) = 2.
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Your solution:
t^2 y’ - 9y = 0 , y(1) = 2
y’ = 9 y / t^2
y’ / y = 9 / t^2
int y’ / y = int 9 / t^2
ln y = -9 / t + c_2
y = e ^ ( -9 / t + c_1)
y = c_1 * e ^ (-9 / t )
2 = c_1 * e ^ -9
c_1 = 2 e ^ 9
Sub this value of c_1 into the equation y = c_1 * e ^ ( -9 / t)
y = 2 e ^ 9 * e ^ ( -9 / t)
y = 2 e ^ ( -9 / t + 9)
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Given Solution:
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Question: 3. (t^2 + t) y' + (2t + 1) y = 0, y(0) = 1.
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Your solution:
t^2 y’ + t y’ + 2 t y + y = 0
After working out the above equation I found it easier to start from the given equation.
(t^2 + t) y' + (2t + 1) y = 0 subtract the y portion
(t^2 + t) y’ = - (2t + 1)y
y’ = - (2t + 1 )y / t^2 + t
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I believe you mean
y’ = - (2t + 1 )y / (t^2 + t)
The signs of grouping here are absolutely essential to avoid ambiguity.
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divide each side by y to separate y and t
y’ / y = -(2t + 1) / t^2 + t
take integrals of each side
int [y’ / y] = int [-(2t + 1) / t^2 + t ]
I do not know how to continue.
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The right-hand side can be easily integrated using the substitution u = t^2 + t.
y ' should be written as dy/dt, in which case your equation would read
dy/y = (-2t+1) / (t^2 + t) dt
It should be clear how the left-hand side would be integrated.
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However this section is on linear equations, and you should be solving these equations as such.
The equation is first-order linear homogeneous, with p(t) = (2 t + 1) / (t^2 + t).
You should solve it accordingly.
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Given Solution:
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Question: 4. y ' + sin(3 t) y = 0, y(0) = 2.
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Your solution:
y’ = -sin(3t) y
y’ / y = -sin(3t)
int [y’ / y] = int [- sin(3t) ]
ln y = 1 / 3 cos(3t) + c_1
y = e^(1 / 3 cos(3t) + c_1)
y = c_1*e^(1 / 3 cos(3t))
Plug in the given y(0) = 2
2 = c_1*e^(1 / 3 cos(0))
2 = c_1*e^(1 / 3)
c_1 = 2 / e^(1/3)
y = 2 * e ^ 1 / 3 (cos(3t) - 1)
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Your solution is correct. However, you need to solve these equations using the methods of this section.
This equation is first-order linear homogeneous with p(t) = sin(3 t).
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Given Solution:
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Question: 5. Match each equation with one of the direction fields shown below, and explain why you chose as you did.
y ' - t^2 y = 0
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y ' - y = 0
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y' - y / t = 0
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y ' - t y = 0
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y ' + t y = 0
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A
y ' - y = 0
The slope y’ is equal to the y value. As the y value increases, so does the slope and vice versa.
B
y ' - t y = 0
The slope y’ is equal to the product of t and y. With negative t and positive y the value of the slope is negative, becoming increasingly negative as we travel away from the origin.
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What you say is also true of graph E.
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As you approach the origin the product yt approaches zero. However if you approach the origin through any straight line other than one of the coordinate axes, this is not the case (this is most obvious for the lines y = t and t = -t).
Along the line y = t, y ' = t y increases with increasing values of y and t. On graph C the values on the line y = t are constant, which eliminates graph C as a possibility.
You'll want to take another look at graph E.
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C
y' - y / t = 0
The slope y’ is equal to y * t. When y = 1, the slope is a constant one. When y is constant, and t increases the slope becomes close and closer to zero.
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The equation y ' - y / t = 0 rearranges to the form
y ' = y / t,
implying slopes which along any horizontal line decrease in magnitude as we move away from the y axis. For horizontal lines above the t axis y is positive, so the slopes will be positive to the right of the y axis and negative to the left.
The only graph which satisfies these criteria is C.
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D
This graph is not featured in the above equations, however at y = 0 the slope y’ is infinity or “straight up” which means here the y’ equation is being divided by a y value of 0.
Example: y’ - ‘something ‘ / y = 0.
E
y ' - t^2 y = 0
The slope y’ is equal to t^2 * y. When y and t are both equal to 1, the slope is 1. When we travel in the positive t direction at an equal y value the slope increases exponentially.
F
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t^2 y is positive when y is positive, negative when y is negative. This eliminates graph F.
y ' = t^2 y implies a direction field in which the magnitudes of the slopes along any line where y is a nozero constant and t is not zero (i.e., along any horizontal line except the t axis), increase in magnitude at an increasing rate as we move away from the y axis. This occurs because the values of t^2 increase at an increasing rate as | t | increases.
This is the case for graphs B and E.
Further confirming the candidacy of both B and E, along any vertical line (on which t is constant) with the exception of the t axis, the magnitude of t^2 y will increase as we move away from the t axis, taking positive values above the t axis and negative values below. The slopes of both graphs graph behave in this manner.
t^2 is always positive, so the slopes will always be positive when y is positive and negative when y is negative. Only graph E has this characteristic.
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6. The graph of y ' + b y = 0 passes through the points (1, 2) and (3, 8). What is the value of b?
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Your solution:
Slope intercept form
y - y_1 = m (x - x_1)
Using the two points given this becomes
8 - 2 = m (3 - 1)
m = 3
Here I am assuming that m, the slope, can be interchanged with y’, also a slope.
If y’ = m = 3 then through plugging in the initial point of (1 , 2)
3 + 2 b = 0
b = - 3 / 2
??? I do not think this is correct because the b value would changed if I used the point (3, 8 ) instead.
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The solution to this equation is not linear, so you can't use the equation of a straight line to model the equation.
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What is the general solution of this equation, and what sort of function should you therefore be using to model the solution?
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confidence rating #$&*:
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Given Solution:
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Question:
7. The equation y ' - y = 2 is first-order linear, but is not homogeneous.
If we let w(t) = y(t) + 2, then:
What is w ' ?
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What is y(t) in terms of w(t)?
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What therefore is the equation y ' - y = 2, written in terms of the function w and its derivative w ' ?
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Now solve the equation and check your solution:
Solve this new equation in terms of w.
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Substitute y + 2 for w and get the solution in terms of y.
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Check to be sure this function is indeed a solution to the equation.
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Your solution:
w(t) = y(t) + 2
w’ = 1
y(t) in terms of w(t) would be
y(t) = w(t) - 2
y’ - y = 2 written in terms of the function w and its derivative w’
y’ - w(t) + 2 = 2
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w ' = (y + 2) ' = y ', not 1.
The equation becomes
w ' = w.
You should complete this problem.
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Given Solution:
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Question: 8. The graph below is a solution of the equation y ' - b y = 0 with initial condition y(0) = y_0. What are the values of y_0 and b?
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Your solution:
??? I do not understand what this question is asking.
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What are the solutions of the given equation?
What is y(0) for the given graph?
Can you then find the value of b?
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Given Solution:
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Not bad overall, but you should give some more thought to some of these problems.
You are invited to resubmit, per the note below, but this won't be a requirement.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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