Query_01

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course MTH-279

06/15 around 8:30PM.

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’ - 2y = 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

confidence rating #$&*:

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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

t^ 2 dy/dt = 9y

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 = 2 e ^ 9

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

confidence rating #$&*:

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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 + 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

cannot divide by zero, so c is undefined by this condition.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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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: 2

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Given Solution:

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Self-critique (if necessary):

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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, because it contains the direction field where the slope changes to zero and then back to what it originated as

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y ' - y = 0

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A, because it contains the direction field where the slope stays the same across the t axis given no influence from a t in the equation

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y' - y / t = 0

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C, since y' = y / t, the slopes will be getting smaller and smaller due to t growing faster than y

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y ' - t y = 0

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B, when finding the slope, y' = t y, in the first quadrant the slopes will be positive and the fourth quadrant negative

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y ' + t y = 0

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F, y' = -t y, in the first quadrant the slopes will be negative and the fourth quadrant negative

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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:

Use the rise / run method to find the slope?

(8-2) / (3-1) = 3 = y' -- plug this into the original

3 = -b y -- using (3,8) to solve for b

3 = -b (8)

b = - 3 / 8

@&
The slope you have found is not an instantaeous slope, but an average slope. You don't know at what point the slope has this value.

The equation y ' + b y = 0 is first-order linear homogeneous, with solution y = A e^(-b t).
The graph passes through (1, 2), so
2 = A e^(- b * 1)
The graph passes through (3, 8), so
8 = A e^(-b * 3).
Thus we have two simultaneous equations in A and b.

Solving these two equations, you will be able to substitute into the form y = A e^(-b t) and verify by substitution that your result is a solution to the equation.
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confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating:

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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(t) = y(t) + 2

w'(t) = (y(t) + 2)' = y'(t) + (2)' = y'(t)

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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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y ' - y = 2

w'(t) - (w(t) - 2) = 2

w'(t) - w(t) + 2 = 2

w'(t) - w(t) = 0

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Now solve the equation and check your solution:

Solve this new equation in terms of w.

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w(t) = c*e^t

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Substitute y + 2 for w and get the solution in terms of y.

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w(t) = c*e^t

y + 2 = c*e^t

y = c*e^t -2

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Check to be sure this function is indeed a solution to the equation.

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y ' - y = 2

c*e^t - (c*e^t -2) = 2

c*e^t - c*e^t + 2 = 2

2 = 2

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Your solution:

Below each respective question

confidence rating #$&*:

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

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Given Solution:

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Self-critique (if necessary):

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(t) = c * e^(-bt)

y_0 = 1 -> y intercept.

y(0) = c * e^(-b*0) = 1

c = 1

y(1) = c * e^(-b*1) = 2

y(1) = e^(-b) = 2

b = -ln(2)

Confidence rating: 2

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Given Solution:

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Self-critique (if necessary):

Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

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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(t) = y(t) + 2

w'(t) = (y(t) + 2)' = y'(t) + (2)' = y'(t)

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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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y ' - y = 2

w'(t) - (w(t) - 2) = 2

w'(t) - w(t) + 2 = 2

w'(t) - w(t) = 0

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Now solve the equation and check your solution:

Solve this new equation in terms of w.

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w(t) = c*e^t

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Substitute y + 2 for w and get the solution in terms of y.

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w(t) = c*e^t

y + 2 = c*e^t

y = c*e^t -2

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Check to be sure this function is indeed a solution to the equation.

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y ' - y = 2

c*e^t - (c*e^t -2) = 2

c*e^t - c*e^t + 2 = 2

2 = 2

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Your solution:

Below each respective question

confidence rating #$&*:

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

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Given Solution:

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Self-critique (if necessary):

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(t) = c * e^(-bt)

y_0 = 1 -> y intercept.

y(0) = c * e^(-b*0) = 1

c = 1

y(1) = c * e^(-b*1) = 2

y(1) = e^(-b) = 2

b = -ln(2)

Confidence rating: 2

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Given Solution:

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Self-critique (if necessary):

Self-critique rating:

Self-critique (if necessary):

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Self-critique rating:

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Query_01

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