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course Mth 279
7:00 PM October 4th
Class Notes and q_a_ for class 110124.________________________________________
This document and the next are supplemented by Chapter 2 of the text.
This should be submitted as a q_a_ document, filling in answers in the usual manner, between the marks
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The **** mark and the #$&* mark should each appear by itself, on its own line.
We show the following:
y ' + t y = 0 has solution y = e^(-t).
If y = e^(-t) then y ' = -t e^(-t) so that
y ' + t y becomes -t e^-t + t e^-t, which is zero.
y ' + sin(t) y = 0 has solution y = e^(cos t)
If y = e^(cos t) then y ' = -sin(t) e^(cos(t)) so that
y ' + t y becomes -sin(t) e^(cos(t)) + sin(t) e^cos(t) = 0
y ' + t^2 y = 0 has solution y = e^(-t^3 / 3)
This is left to you.
If y= e^(-t^3 / 3) then y=-t^3/3 e^(-t^3 / 3)
So that y+ty becomes -t^3/3 e^(-t^3 / 3)+ t^3/3 e^(-t^3 / 3)=0
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What do all three solutions have in common?
Some of this is left to you.
However for one thing, note that they all involve the fact that the derivative of a function of form e^(-p(t)) is equal to -p'(t) e^(-p(t)).
And all of these equations are of the form y ' + p(t) y = 0.
Now you are asked to explain the connection.
All of these equations are of a form which when the derivative of their solution is added they become equal to zero.
@& That is true of the solution of every differential equation.
The expressions we use here have a pattern unique to this type of equation.*@
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What would be a solution to each of the following:
y ' - sqrt(t) y = 0?
If we integrate sqrt(t) we get 2/3 t^(3/2).
The derivative of e^( 2/3 t^(3/2) ) is t^(1/2) e^ ( 2/3 t^(3/2) ), or sqrt(t) e^( 2/3 t^(3/2) ).
Now, if we substitute y = sqrt(t) e^( 2/3 t^(3/2) ) into the equation, do we get a solution? If not, how can we modify our y function to obtain a solution?
Not sure how to answer this one. Confused to where the 2/3t^(2/3) came from in the first place.
@& The question is whether
y = sqrt(t) e^( 2/3 t^(3/2) )
is a solution to the given equation.
A secondary question is where that 2/3 t^(3/2) came from, and if you don't discover that yourself it will be pointed out later, but the more important thing is to get some insight by answering the main question.*@
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sqrt(t) y ' + y = 0?
The rest of our equations started with y ' . This one starts with sqrt(t) y '.
We can make it like the others if we divide both sides by sqrt(t).
We get
y ' + 1/sqrt(t) * y = 0.
Follow the process we used before.
We first integrated something. What was it we integrated?
1/sqrt(t)
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We then formed an exponential function, based on our integral. That was our y function. What y function do we get if we imitate the previous problem?
y=1/sqrt(t)e^(2/3t^(2/3))
@& Before we integrated sqrt(t) to get 2/3 t^(3/2).
This time 1/sqrt(t) is the coefficient of y, so if we imitate the procedure from the preceding we will integrate 1 / sqrt(t), which gives us something different (and does not involve the 2/3 power).*@
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What do we get if we plug our y function into the equation? Do we get a solution? If not, how can we modify our y function to obtain a solution?
2sqrt(t)+ 1/sqrt(t)e^(2/3t^(2/3)) = 0
No solution not sure what modification would obtain a solution. Is this the correct equation?
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t y ' = y?
If we divide both sides by t and subtract the right-hand side from both sides what equation do we get?
?
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Why would we want to have done this?
?
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Imitating the reasoning we have seen, what is our y function?
?
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Does it work?
?
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y ' + p(t) y = 0 has solution y = e^(- int(p(t) dt)).
This says that you integrate the p(t) function and use it to form your solution y = e^(- int(p(t) dt)).
Does this encapsulate the method we have been using?
Yes
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Will it always work?
Yes, because the next step is to differentiate the p(t) function and add it to the -p(t) function which will equal zero.
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What do you get if you plug y = e^(-int(p(t) dt) into the equation y ' + p(t) y = 0?
Zero?
@& Ultimately you will get 0, but you need to show the intermediate steps.
For this y function, what is y ' ?*@
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Is the equation satisfied?
Yes?
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y ' + p(t) y = 0 is the general form of what we call a first-order linear homogeneous equation. If it can be put into this form, then it is a first-order linear homogeneous equation.
Which of the following is a homogeneous first-order linear equation?
y * y ' + sin(t) y = 0
We need y ' to have coefficient 1. We get that if we divide both sides by y.
Having done this, is our equation in the form y ' + p(t) y = 0?
No, dividing by y will remove the y term from the p(t)y portion
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Is our equation therefore a homogeneous first-order linear equation?
No
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t * y ' + t^2 y = 0
Once more, we need y ' to have coefficient 1.
What is your conclusion?
t * y ' + t^2 y = 0 divided by t equals y+ty=0
Therefore, this is a homogeneous first-order linear equation.
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cos(t) y ' = - sin(t) y
Again you need y ' to have coefficient 1.
Then you need the right-hand side to be 0.
Put the equation into this form, then see what you think.
cos(t) y ' = - sin(t) y divided by cos(t) equals y=-tan(t)y
Adding tan(t)y to both sides equals y+tan(t)y=0
Therefore, this is a homogeneous first-order linear equation.
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y ' + t y^2 = 0
What do you think?
The y term of p(t)y means this is not a homogeneous first-order linear differential equation.
@& Right. y^2 is not linear.*@
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y ' + y = t
How about this one?
There is no way to make the y and t term of the form p(t)y to make this a homogeneous first-order linear differential equation.
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Solve the equations above that are homogeneous first-order linear equations.
t * y ' + t^2 y = 0 divided by t equals y+ty=0
y+tan(t)y=0
multiply both sides by e^(-logcos(t)), results in the derivative with respect to t of e^(-logcos(t))*y which is ysin(t)cos^(-y-1)*(t)
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Verify the following:
If you multiply both sides of the equation y ' + t y by e^(t^2 / 2), the result is the derivative with respect to t of e^(t^2 / 2) * y.
The derivative with respect to t of e^(t^2 / 2) * y is easily found by the product rule to be
(e^(t^2 / 2) * y) '
= (e ^ (t^2 / 2) ) ' y + e^(t^2/2) * y '
= t e^(t^2/2) * y + e^(t^2 / 2) * y '.
If you multiply both sides of y ' + t y by e^(t^2 / 2) you get e^(t^2 / 2) y ' + t e^(t^2 / 2).
Same thing.
Now, what is it in the original expression y ' + t y that led us to come up with t^2 / 2 to put into that exponent?
The integral of t which was the p(t) term?
@& Right.*@
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If you multiply the expression y ' + cos(t) y by e^(sin(t) ), the result is the derivative with respect to t of e^(-sin(t)) * y.
Just do what it says. Find the t derivative of e^(sin(t) ) * y. Then multiply both sides of the expression y ' + cos(t) y by e^(sin(t) * y).
Derivative of e^(sin(t) )*y=y*e^(sin(t))*cos(t) multiply both sides by e^(sin(t)*y)
=
@& e^(sin(t) )*y is not y*e^(sin(t))*cos(t).
You need to apply the product rule.
(e(sin(t) * y) = (e^(sin(t)) ' * y + e^(sin(t) (y) '.
If you multiply the expression y ' + cos(t) y by e^(sin(t) ), what do you get?
The result should be equal to your derivative.*@
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How did we get e^(sin(t)) from of the expression y ' + cos(t) y? Where did that sin(t) come from?
The integral of cos(t) which was the p(t) term?
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If you multiply both sides of the equation y ' + t y = t by e^(t^2 / 2), the integral with respect to t of the left-hand side will be e^(t^2 / 2) * y.
You should have the pattern by now. What do you get, and how did we get t^2 / 2 from the expression y ' + t y = t in the first place?
It is the integral of t which is the p(t) term?
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The equation becomes e^(t^2 / 2) * y ' + t e^(t^2 / 2) y = t e^(t^2 / 2).
The left-hand side, as we can easily see, is the derivative with respect to t of e^(t^2 / 2) * y.
So if we integrate the left-hand side with respect to t, since the left-hand side is the derivative of e^(t^2 / 2) * y, an antiderivative is e^(t^2 / 2) * y.
Explain why it's so.
Confused here
@& There are multiple statement in the preceding. I need you to be more specific about the point(s) of confusion.*@
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Having integrated the left-hand side, we integrate the right-hand side t e^(t^2 / 2).
What do you get? Be sure to include an integration constant.
=e^(t^2/2)+c
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Set the results of the two integrations equal and solve for y. What is your result?
e^(t^2 / 2) * y =e^(t^2/2)+c
y=c
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Is it a solution to the original equation?
Should I be able to solve for a value of c to check if this is a solution of original equation??
@& The integration constant should remain unspecified.
You need to solve that equation for y.*@
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If you multiply both sides of the equation y ' + p(t) y = g(t) by the e raised to the t integral of p(t), the left-hand side becomes the derivative with respect to t of e^(integral(p(t) dt) ) * y.
See if you can prove this.
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@& I've inserted some questions and some guidance.
This document demonstrates why it's so, but the bottom line is that you solve the homogeneous equation
y ' + p(t) y = 0
and get solution y = e^(- integral(p(t) )
You solve the nonhomogeneous equation
y ' + p(t) y = 1
by first multiplying both sides by e^(integra(p(t)), then observing that the left-hand side is a perfect differential, which allows you to integrate both sides and then solve for y.
A revision isn't absolutely required, but you should consider revising at least some of your work in accordance with my notes. You appear to understand enough to go on, so this is optional. Weight the time required vs. the potential benefit.
&#Please see my notes and, unless my notes indicate that revision is optional, submit a copy of this document with revisions 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.
If my notes indicate that revision is optional, use your own judgement as to whether a revision will benefit you. &#
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