Query 01

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course Mth 279

1/1/14 around 4pm

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

dy/dt - 2y = 0

dy/dt = 2y

dy = 2y dt

dy/y = 2 dt

int(dy/y) = 2 * int(dt)

ln(y) = 2t + c

y = e^(2t + c)

y = c * e^(2t)

y(1) = 3

y(1) = c * e^(2(1)) = 3

c = 3/e^2

y = (3/e^2) * e^(2t)

y = 3e^(2t) /e^2

@&
You solved the equation using separation of variables. Your solution is correct, but the corresponding section is on the topic of linear equations, and the equation should have been solved accordingly.

Be sure you know how to solve this as a linear equation.
*@

confidence rating #$&*:8232;

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:

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Question: 2. t^2 y ' - 9 y = 0, y(1) = 2.

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Your solution:

t^2 y ' - 9 y = 0

t^2 dy/dt = 9 y

dy/y = 9 dt/t^2

int (dy/y) = 9 * int (dt/t^2)

ln(y) = -9/t + c

y(t) = e^(c - 9/t)

y(t) = c/e^(9/t)

y(1) = 2

y(1) = c/e^(9/1) = 2

c = 2e^9

y(t) = 2e^9 / e^(9/t)

confidence rating #$&*:8232;

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:3

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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 + t) y' + (2t + 1) y = 0

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

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

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

ln(y) = -ln(t^2 + t) + c

y(t) = c/(t^2 + t)

y(0) = 1

y(0) = c/(0^2 + 0) = c/0 = 1

C is undefined due to not being able to divide by zero

confidence rating #$&*:8232;

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution: OK

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Self-critique (if necessary):3

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Self-critique rating:3

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Question: 4. y ' + sin(3 t) y = 0, y(0) = 2.

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Your solution:

y ' + sin(3 t) y = 0

dy/dt + sin(3 t) y = 0

dy/dt = -sin(3 t) dt

int (dy/dt) = - int (sin(3 t) dt)

ln(y) = cos(3t) /3 + c

y(t) = e^(cos(3t) /3 + c)

y(t) = c * e^(cos(3t) /3)

y(0) = 2

y(0) = c * e^(cos(3* 0) /3) = 2

y(0) = c * e^(1/3) = 2

c = 2/e^(1/3)

y(t) = 2/e^(1/3) * e^(cos(3t) /3)

y(t) = 2e^(cos(3t) /3) /e^(1/3)

confidence rating #$&*:8232;

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:3

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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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E, the only direction where slope changes to zero and goes back to where it began.

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@&
This equation corresponds to graph B.
y ' = t^2 y so the magnitude of the slope increases with distance from y axis (i.e., with increasing | t | ) and with distance from x axis (i.e., with increasing | y | ); slope is negative when y is negative

*@

y ' - y = 0

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A, only direction field where the slope stays the same

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y' - y / t = 0

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Right.

This equation corresponds to graph A; y ' = y so slope is positive above the x axis, negative below; the magnitude of the slope increases with distance from x axis and is the same on any horizontal line.
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C, y’=y/t, the slopes will begin to get smaller and smaller due to a growing t

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y ' - t y = 0

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Right.

y ' = y / t; along any vertical line slope is equal to y, so magnitude of the slope increases as you move away from the x axis along such a line.

The slope increases in magnitude as you approach the y axis.

The slope is positive in the first and third quadrants, negative in the second and fourth.

Only graph C has these characteristics.

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B, it is in the first quadrant and the forth will be negative

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y ' + t y = 0

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y ' = - t y. The slope is negative in the first and third quadrants, positive in the second and fourth.

The magnitude of the slope increases as you move away from the x axis along any vertical line, and as you move away from the y axis along any horizontal line.

Graph F has these characteristics.

*@

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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:

Im not sure how to go about this one

@&
The general solution to this equation is

y = A e^(-b t) .

The coordinates of each point must satisfy this equation. So each point gives you an equation with specific values of t and y, with A and b as unknowns.

So you have two simultaneous equations in the two unknowns A and b.
*@

confidence rating #$&*:8232;

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Given Solution:

&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&

Self-critique (if necessary): I know how to find the average slope but I’m not sure how that will help since it yields an exponential equation

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Self-critique rating:2

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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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w' = (2+y) + 2 We get 2 + y from y' = 4 + y

@&
If w = y + 2, then w ' = y '.

Thus w ' - y = 2, so w ' = y + 2, and y + 2 = w.

So with the change of variable w = y + 2, the equation is

w ' = w
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What is y(t) in terms of w(t)?

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y(t)=w(t) -2

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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: Not to sure on this one

@&
See if the start I gave you above helps you get further on this problem.
*@

confidence rating #$&*:8232;

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Given Solution:

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Self-critique (if necessary):NONE

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Self-critique rating:

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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:

y(0) = y_0 is equal to 1 which is the C value

approximate with (-1, 0.5)

y = e^(-b t)

0.5 = e^(-b (-1))

0.5 = e^(b) —> natural log

ln(0.5) = ln(e^(b))

b = - 0.69

@&
Good.
*@

confidence rating #$&*:8232;

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:3

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

@&
Check my first note, which applies to a number of problems, and be sure you can solve these problems by the methods of the section, which is on linear homogeneous equations.
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Query 01

@& You solved the equation using separation of variables. Your solution is correct, but the corresponding section is on the topic of linear equations, and the equation should have been solved accordingly. Be sure you know how to solve this as a linear equation. *@

@& This equation corresponds to graph B. y ' = t^2 y so the magnitude of the slope increases with distance from y axis (i.e., with increasing | t | ) and with distance from x axis (i.e., with increasing | y | ); slope is negative when y is negative *@

@& Right. This equation corresponds to graph A; y ' = y so slope is positive above the x axis, negative below; the magnitude of the slope increases with distance from x axis and is the same on any horizontal line. *@

@& The general solution to this equation is y = A e^(-b t) . The coordinates of each point must satisfy this equation. So each point gives you an equation with specific values of t and y, with A and b as unknowns. So you have two simultaneous equations in the two unknowns A and b. *@

@& If w = y + 2, then w ' = y '. Thus w ' - y = 2, so w ' = y + 2, and y + 2 = w. So with the change of variable w = y + 2, the equation is w ' = w *@

@& See if the start I gave you above helps you get further on this problem. *@

@& Good. *@

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

@& Check my first note, which applies to a number of problems, and be sure you can solve these problems by the methods of the section, which is on linear homogeneous equations. *@

@&
You solved the equation using separation of variables. Your solution is correct, but the corresponding section is on the topic of linear equations, and the equation should have been solved accordingly.

Be sure you know how to solve this as a linear equation.
*@

@&
This equation corresponds to graph B.
y ' = t^2 y so the magnitude of the slope increases with distance from y axis (i.e., with increasing | t | ) and with distance from x axis (i.e., with increasing | y | ); slope is negative when y is negative

*@

@&
Right.

This equation corresponds to graph A; y ' = y so slope is positive above the x axis, negative below; the magnitude of the slope increases with distance from x axis and is the same on any horizontal line.
*@

@&
The general solution to this equation is

y = A e^(-b t) .

The coordinates of each point must satisfy this equation. So each point gives you an equation with specific values of t and y, with A and b as unknowns.

So you have two simultaneous equations in the two unknowns A and b.
*@

@&
If w = y + 2, then w ' = y '.

Thus w ' - y = 2, so w ' = y + 2, and y + 2 = w.

So with the change of variable w = y + 2, the equation is

w ' = w
*@

@&
See if the start I gave you above helps you get further on this problem.
*@

@&
Good.
*@

@&
Check my first note, which applies to a number of problems, and be sure you can solve these problems by the methods of the section, which is on linear homogeneous equations.
*@