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course Mth 279
Jan 29 around 11:45pm. I got hung up on a couple, but attempted what I could!
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.
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y’ + t^2*y =0
y=e^(-t^3/3)
y’=-t^2*e(-t^3/3)
(-t^2*e^(-t^3/3)) + (t^2*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.
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They all start the expression out with the negative value of the coefficient in front of “t”, followed by the positive which cancels out and equals 0.
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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?
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No, we’re looking for –Int (p(t) dt) so y(t)=e^(-Int(p(t) dt)).
-Int(sqrt (t))=(2/3t^(3/2)) so y’= sqrt(t) e^(2/3 t^(3/2))
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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?
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Y’ + (1/sqrt t)*y =0
P(t)=(1/sqrt t)
-Int (p(t) dt)
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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?
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Y(t)= e^(-Int (p(t) dt))
Y(t)= e^(-Int((1/sqrt) t dt)
=e^(-2*sqrt t)
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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?
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Find y’.
Y= e^(-2*sqrt t)
Y’= (-e^(-2*sqrt t)/sqrt t)
YES, you get a solution.
(-e^(-2*sqrt t)/sqrt t) + (1/sqrt t)*(e^-2*sqrt t) =0
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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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Y’=y/t
Y’-(y/t)=0
Y’-(1/t)y=0
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Why would we want to have done this?
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To get it in the form (y’ + ty=0) or (y’ +p(t)y=0)
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Imitating the reasoning we have seen, what is our y function?
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Y=e^(Int -1/t dt)
=e^(-ln (abs t))
=abs (1/t)
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Does it work?
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Y’= (- abs (1/t))/t
(- abs (1/t))/t – (1/t)(1/t) =0
?!
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@& If t > 0 then the derivative of 1/t is -1/t^2.*@
• 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?
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YES
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Will it always work?
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YES?!
A solution of y’=-p(t)y
Y=e^(-Int (p(t) dt))
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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?
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I’m having trouble on this one…
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@& The derivative of int(p(t) dt) is p(t), since the integral of p(t) is int(p(t) dt).
So the derivative of e^(-int(p(t) dt) is - p(t), whatever p(t) is.*@
Is the equation satisfied?
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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?
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If you divide all sides by “y” then:
Y’+ sin(t)=0
The “y” canceled out.
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@& Right.
It follows that the equation is not of the form y ' + p(t) y = 0.*@
Is our equation therefore a homogeneous first-order linear equation?
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NO
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• t * y ' + t^2 y = 0
Once more, we need y ' to have coefficient t.
What is your conclusion?
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Y’ + ty =0
1st order linear homogeneous eq.
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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.
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Y’ + tan(t) y=0
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• y ' + t y^2 = 0
What do you think?
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If you divide all sides by “y”, then there wouldn’t be a coefficient of 1 in front of y’….
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• y ' + y = t
How about this one?
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Y’ + y-t =0
NO
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Solve the problems above that are homogeneous first-order linear equations.
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?
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Since p(t)=t
=e^(-Int (p(t) dt))
=e^(-Int (t dt))
=e^(-t^2/2)
Wouldn’t (t^2/2) be negative?!
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@& If you multiply both sides by e^(-t^2 / 2) then the derivative of e^(-t^2 / 2) * y won't be the left-hand side of the equation. *@
• If you multiply both sides of the equation 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).
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Use product rule:
=(e^(sin t))*(y’) + (y)*(e^(sin t))’
=[(e^sin t)*y] + [cos(t)*e^(sin t)*y]
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@& I think you mean
=[(e^sin t) * y ' ] + [cos(t)*e^(sin t)*y],
which would be correct.*@
How did we get e^(sin(t)) out of the expression y ' + cos(t)? Where did that sin(t) come from?
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Dividing both sides by (e^(sin t)), cancels out.
Int p(t) dt
Int (cos t dt)
=e^(sin t)
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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 do we get t^2 / 2 from the expression y ' + t y = t in the first place?
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You get (t^2/2) by:
Since p(t)=t
Int (t dt)=(t^2/2)
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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.
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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.
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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?
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[(e^(t^2/2))*y + c1] = [(e^(t^2/2)) + c2]
Y=c2-c1
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@& The solution of [(e^(t^2/2))*y + c1] = [(e^(t^2/2)) + c2] for y is
y = = [(e^(t^2/2) + c2 - c1) / e^(t^2 / 2)
which is equal to
1 + (c2 - c1) / e^(t^2 / 2).
c1 and c2 are arbitrary constants, so c2 - c1 can be expressed in terms of a single arbitrary constant c as
1 + c / (e^(t^2 / 2)
or as
1 + c e^-(t^2 / 2)
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Is it a solution to the original equation?
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I’m not sure if I did everything right since y equals a constant…
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• If you multiply both sides of the equation y ' + p(t) y = g(t) by 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 do this.
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@& Good, for the most part.
See my notes and let me know if you have additional questions. *@