query_01

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course MTH 279

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:

The first order linear homogeneous equation is in the form y' + p(t)y = g(t). To solve the IVP we will form the general solution y = Ce^(-P(t)).

p(t) = -2t. An antiderivative is P(t) = (integral of) -2t dt = -t^2. Having an antiderivative P(t), we obtain the general solution y = Ce^(t^2).

Using the initial condition y(1) = 3, we have Ce^((1)^2) = 3, or Ce = 3, making C = 3/e

The unique solution of the IVP is y = (3/e)e^(t^2)

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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:

First we must divide by t^2 to put the equation in standard form. So our equation is rewritten y' - 9y/(t^2) = 0.

p(t) = 9/(t^2). A convenient antiderivative is P(t) = (integral of) 9/(t^2) dt = 9 ln |t^2| = -9/t.

Having an antiderivative P(t), we obtain the general solution: y = Ce^-(-9/t) = Ce^9/t. Using the initial condition y(1) = 2, Ce^9/1 = 2, we know that C = 2/e^9.

A unique solution exists: y = (2/e^9)e^-(-9/t).

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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:

In order to put the equation in standard form, we must divide by (t^2+t). Our equation is rewritten to be y' + (2t + 1) y/(t^2 + t) = 0.

p(t) = (2t+1)/(t^2+t). A convenient antiderivative is P(t) = (integral of) (2t+1)/(t^2+t) dt = log(t) + log(t + 1).

Having P(t), we obtain the general solution: y = Ce^-(log(t) + log(t + 1)) = C(log(t) - log(t+1)). Using the initial condition, we calcualte

C(log(0)-log((0)+1) = 1. Therefore the answer is C can be anything other than 0.

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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:

p(t) = sin(3t).

P(t) = (integral of) sin(3t) = -1/3 cos(3x).

y = Ce^-(-1/3 cos(3x)) = Ce^1/3 cos(3x). Using the initial condition y(0) = 2, we plug in to get

=Ce^1/3cos(3(0)) = 2

=Ce^1/3cos(0) = 2

=Ce^1/3 = 2

so C = 2 / (e^1/3).

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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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Rewrite y ' - t^2 y = 0 as y' = t^2y.

Then using a graph, I get the following chart

t y y'

0 0 0

1 1 1

-1 1 1

1 -1 -1

-1 0 0

0 -1 0

1 0 0

0 1 0

2 -2 -8

-2 2 8

2 2 8

-2 -2 -8

-1 -1 -1

We know that the answer is E.

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y ' - y = 0

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y' - y = 0 is not dependent on t. Calculating the values for y' and y, shown in the chart below,

y' y

-2 -2

-1 -1

0 0

1 1

2 2

We know that the answer is A.

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y' - y / t = 0

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Calculating the value for y' = y/t for t and y, I get the following chart:

t y y' = y/t

1 1 1

-1 -1 -1

1 -1 -1

0 0 0

1 0 0

-1 0 0

2 2 1

-2 -2 1

-2 2 -1

2 -2 -1

We know that the answer is C.

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y ' - t y = 0

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Rewriting y' - ty = 0 as y' = ty, we can use the equation to calculate the following chart:

t y y'

0 0 0

1 1 1

-1 1 -1

1 -1 -1

-1 -1 1

-2 -2 4

-2 2 -4

2 -2 -4

2 2 4

0 1 0

1 0 0

We know that the answer is D.

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y ' + t y = 0

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Rewriting the equation as y' = -ty, we can use the equation to calculate the chart:

t y y'

0 0 0

1 1 -1

-1 1 1

1 -1 1

0 -1 0

0 1 0

1 0 0

-1 0 0

2 -2 4

-2 2 4

2 2 -4

-2 -2 -4

We know that the answer is F.

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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:

We know that the solution of y' = -by is y = -b(y^2)/2.

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The solution would be y = A e^(-b t).

Thus

2 = A e^(-b * 1) and
8 = A e^(-b * 8).

Dividing the first equation by the second we get

1/4 = e^(7 b),

which can easily be solved for b.
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Given Solution:

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Self-critique (if necessary): I'm very confused about how to approach this problem.

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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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If w(t) = y(t) + 2 = y'(t), so w'(t) = y''(t) = 2 + y'(t)

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What is y(t) in terms of w(t)?

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In terms of w(t), y(t) would equal y'(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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w - (integral of) w = 2

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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 - (integral of) w = 2

w - (w^2)/2 = 2

w = 2 + (w^2)/2

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

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

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

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Your solution:

If w(t) = y(t) + 2 = y'(t), so w'(t) = y''(t) = 2 + y'(t)

In terms of w(t), y(t) would equal y'(t).

w - (integral of) w = 2

w - (w^2)/2 = 2

w = 2 + (w^2)/2

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

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Given Solution:

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Self-critique (if necessary):

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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' - by = 0 can be rewritten to be y' = by. Then we can take the (integral of) by dy = b(y^2)/2.

Imposing the initial condition, we know that y(0) = y_0, so b(0^2)/2 = y_0. So y(0) = 0.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):

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Self-critique rating: "

query_01

@& The solution would be y = A e^(-b t). Thus 2 = A e^(-b * 1) and 8 = A e^(-b * 8). Dividing the first equation by the second we get 1/4 = e^(7 b), which can easily be solved for b. *@

@& Good work overall. Check solutions at http://vhmthphy.vhcc.edu/tests/shell_01/differential_equations/query_01_solnerngkf0jknj22lkkkls.htm *@

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The solution would be y = A e^(-b t).

Thus

2 = A e^(-b * 1) and
8 = A e^(-b * 8).

Dividing the first equation by the second we get

1/4 = e^(7 b),

which can easily be solved for b.
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@&
Good work overall.

Check solutions at

http://vhmthphy.vhcc.edu/tests/shell_01/differential_equations/query_01_solnerngkf0jknj22lkkkls.htm

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