query 05 redo

#$&*

course Mth 279

7/1 11:07 pm

#$&* course Mth 279

Query 05 Differential Equations*********************************************

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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:

divide by e^t

e(-y) y' = -t + sin(t)

Integrate

-e^(-y) y = -1/2 t^2 - cos(t) + C

The derivative of

-1/2 t^2 - cos(t)

is -t + sin(t), but the derivative of

-e^(-y) y

is (e^(-y) * y - e^(-y) ) y ', not just e^(-y) y '.

Y = 1/2 e^(y) * t^2 + e^y cos(t) - Ce^(y)

Initial conditions

0 = 1/2(0) + e^(y) - C e(y)

0 = e^y (1-C)

C = 1

y = 1/2 e^(y) t^2 + e^(y) cos(t) - e^(y)

####

I understand what you're saying, and I should be using that as a self-check method.

I tried typing this into google and it didn't give me an answer.

The next thing I thought of was integration by parts

e^(-y) y' = sint - t

integrate, u = e^(-y), du = -e^(-y), dv = y', v = y

uv - integral[du*v]

ye^(-y) - integral [ -e^(-y) * y]

I think this is where I am going wrong. I think I do the last term using integration by parts again

u = e^(-y), du = -e^(-y), dv = y, v= 1/2 y^2

y e^(-y) + 1/2 e^(-y) y^2 - integral [-1/2 e^(-y) y^2 ]

Now I have another term that I'd like to integrate by parts, but I'm going to keep getting y to higher powers.

???

I thought I could use u substitution, but its y' not y.

Based off of the hint from the next question:

integral [ e^(-y) y' ] should equal -e^(-y) ?

So then,

-e^(-y) = -cos(t) - 1/2 t^2 + C

e^(-y) = cos(t) + 1/2 t^2 + C

-y = ln(cos(t) + 1/2 t^2 + C)

y = - ln (cos(t) + 1/2 t^2 + C )

0 = -ln (1 + 0 + C)

0 = -ln(1) *ln(C)

C = 0

y = -ln( cos(t) + 1/2 t^2 )

######

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: 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' = 1- 2t

integrate

y^3 y = t - t^2 + C

Your left-hand side is not correct.

The equation becomes

3 y^2 dy = (1 - 2t) dt

An antiderivative of the left-hand side is y^3.

The derivative of y^3 with respect to t is 3 y^2 y '.

####

I think here is where I don't understand. how does integral[ y^3] = 3y^2 y' ???

Every other time doesn't it equal 3y^2 ? ??

And this is called implicit... something?

y^3 = t - t^2 + C

y = cuberoot[ t- t^2 + C]

How do you notate cube roots???

-1 = cuberoot [ 0 - 0 + C]

C = -1

y = cuberoot[ t- t^2 -1]

####

y = (t-t^2 + C)/y^3

initial conditions

-1 = (0 + 0 + C)/y^3

C =- y^3

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

confidence rating #$&*:

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Given Solution:

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

Self-critique (if necessary):

not 100% sure if I did that right. I tried following the example in the book

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Self-critique rating:

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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:

take the derivative(?)

3y^2 y' + 2t + cos(y) y' = 0

y' (3y^2 + cos(y) ) = 2t

The equation

y' (3y^2 + cos(y) ) = 2t

means

(3 y^2 + cos(y) ) dy/dt = 2 t,

which is effectively rearranged to

(3 y^2 + cos(y) ) dy = 2 t dt

@&
This is a good equation, matching the given equation.

*@

and then integrated.

when t = 4_0

y_0 ( cos(y)) = 8

I'm not sure how to go from here. I think I need a refresher on implicit differentiation too.

Do see the Calculus qa review exercises.

####

Do you mean the 12 question qa assignment that was part of orientation??? If so, I did it and received your feedback. I also fixed errors I had in qa 00 that you had given me feedback on with regards to integration. I think I'm going to google integration exercises, although I think I specifically need help with implicit and integration by parts

@&
I remembered incorrectly about posting review suggestions on your Assignments Page. They are posted for the Vector Calculus class but not for this one.

Here's a link with more information. A link to this document is also now included on your Assignments Page.

http://vhcc2.vhcc.edu/dsmith/geninfo/qa_query_etc/differential_equations/review_references_first_semester_calculus.htm
*@

I think I was following an example in the book because I'm not sure how I got to

3y^2 y' + 2t + cos(y) y' = 0

But, I know that I arranged both sides of the equation and took the derivative

3y^2 y' + cos(y) y' = -2t

Why would the next step be integration???

3y + sin(y) = t^2 + C

This is right where we started. ???

####

confidence rating #$&*:

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Given Solution:

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

Self-critique (if necessary):

was I on kind of on the right track ???

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Question: 3.2.10. Solve 3 y^2 y ' + 2 t = 1 with initial condition y(0) = -1.

YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY

Your solution:

3y^2 y' = 1- 2t

integrate

y^3 y = t - t^2 + C

Your left-hand side is not correct.

The equation becomes

3 y^2 dy = (1 - 2t) dt

An antiderivative of the left-hand side is y^3.

The derivative of y^3 with respect to t is 3 y^2 y '.

####

I think here is where I don't understand. how does integral[ y^3] = 3y^2 y' ???

@&
It's the derivative of y^3 that equals 3 y^2 y '.

See the review exercise on Implicit Differentiation, which explains how this arises from the chain rule.
*@

Every other time doesn't it equal 3y^2 ? ??

@&
The derivative of y^3 with respect to y is 3 y^2.

The derivative of y^3 with respect to t is 3 y^2 dy/dt, or 3 y^2 y ', where the ' indicates derivative with respect to t.
*@

And this is called implicit... something?

y^3 = t - t^2 + C

y = cuberoot[ t- t^2 + C]

How do you notate cube roots???

@&
The cube root of a quantity is its 1/3 power.
*@

-1 = cuberoot [ 0 - 0 + C]

C = -1

y = cuberoot[ t- t^2 -1]

####

y = (t-t^2 + C)/y^3

initial conditions

-1 = (0 + 0 + C)/y^3

C =- y^3

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

confidence rating #$&*:

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

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Given Solution:

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

Self-critique (if necessary):

not 100% sure if I did that right. I tried following the example in the book

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Self-critique rating:

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Question: 3.2.24. Solve the equation y ' = (y^2 + 2 y + 1) sin(t) fand determine the t interval over which he solution exists.

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Your solution:

y' / (y^2 + 2y + 1) = sin(t)

integrate

I don't know how to integrate 1/(y^2 + 2y + 1)

I thought I could do u sub, but then I'd get u = y^2 + 2y + 1 and du = 2y +2 and that's not helpful

then I thought of the power rule

(y^2 + 2y + 1)^(-1)

but then I'd end up with ln (y^2 + 2y + 1) / (2y+2)

how is this done???

You made a couple of good tries.

Factor the denominator, which gives you a perfect square. Then do what should be an obvious u substitution.

In general, had the quadratic not factored into a perfect square, you would try to factor and use partial fractions, or alternatively complete the square. These are standard first-year calculus techniques, and you should review them.

- y / (y+1) = -cos(t)

y = cos(t) * (y +1)

solution exists from (- infin., + infin.)

####

y' = (y+1)^2 sin(t)

y' (1/(y+1)^2) = sin(t)

integrate

-1/(y+1) = -cos(t) + C

y+ 1 = sec(t) + C

y = sec(t) -1 + C

The equation exists everywhere except when cos(t) = 0

@&
Good.
*@

####

confidence rating #$&*:

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

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Given Solution:

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Self-critique (if necessary):

just not sure about integrating still

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Self-critique rating:

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Question: 3.2.28. Match the graphs of the solution curves with the equations y ' = - y^2, y ' = y^3 an dy ' = y ( 4 - y).

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Your solution:

For me this is problem 30 in section 2.7

I'm not at all sure how to solve for this problem.

Just by looking at the graphs I think y' = y^3 is graph B, only because it looks like a x^3 equation. By the same reasoning I think y' = -y^2 is graph A and y' = y(4-y) is graph C

I should be able to solve these and then match with the graphs

y' = y^3

integrate

y = 1/4 y^4

But that doesn't look like any of these graphs

I don't think I fully understand this question

???

y' = -y^2

integrate

y = -1/3 y^3, this sorta looks like graph B, but it's facing the wrong way

y' = 4y - y^2

integrate

y = 2y^2 -1/3 y^3, this doesn't really look like any of the graphs either.

According to the graphs in the book:

y'=-y^2 --> Graph C

Concave up, asymptote at t = -1 and approaching 0 as t-> inf

y'= y^3 --> Graph A

Concave up but flipped of A with y --> 0 as t --> -inf and an asymptote at t= 1/2

y'= y(4-y) --> Graph B

For one, by exclusion but also because slope of y'-->0 as y--> 4 and y' --> 0 as y --> -inf

####

I don't understand what you did here. The graph is -y^2 is an upside down parabola, which is concave down. It also does not have an asymptote or approach 0 as t goes to infinity.

y^3 also does not have an asymptote.

If I make a slope segment diagram for all of these, the slope does not depend on the variable t. Therefore the slopes will be the same across each horizontal line.

What exactly are you graphing???

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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:

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

Self-critique (if necessary):

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Self-critique rating:

Check my notes.

You are using appropriate techniques for solving the equations, but as I believe you recognize you are running into problems with integration.

The only remedy is practice on integration. There are numerous web resources.

added notes with ####"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

Check my notes.

You are using appropriate techniques for solving the equations, but as I believe you recognize you are running into problems with integration.

The only remedy is practice on integration. There are numerous web resources.

added notes with ####"

Self-critique (if necessary):

------------------------------------------------

Self-critique rating:

#*&!

query 05 redo

@& This is a good equation, matching the given equation. *@

@& It's the derivative of y^3 that equals 3 y^2 y '. See the review exercise on Implicit Differentiation, which explains how this arises from the chain rule. *@

@& The derivative of y^3 with respect to y is 3 y^2. The derivative of y^3 with respect to t is 3 y^2 dy/dt, or 3 y^2 y ', where the ' indicates derivative with respect to t. *@

@& The cube root of a quantity is its 1/3 power. *@

@& Good. *@

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

@&
This is a good equation, matching the given equation.

*@

@&
It's the derivative of y^3 that equals 3 y^2 y '.

See the review exercise on Implicit Differentiation, which explains how this arises from the chain rule.
*@

@&
The derivative of y^3 with respect to y is 3 y^2.

The derivative of y^3 with respect to t is 3 y^2 dy/dt, or 3 y^2 y ', where the ' indicates derivative with respect to t.
*@

@&
The cube root of a quantity is its 1/3 power.
*@

@&
Good.
*@