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course MTH 279
Note: I accidentally sent a blank submit with only one access code typed in just before this one. My autofill messed me up on that one. " "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 e^(-t^3 / 3)
-t^2 e^(-t^3 / 3)+t^2 e^(-t^3 / 3)=0
0=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 are formed from the exponent of the integral of p(t). Other than that and the observation made above, I'm not sure.
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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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The sqrt cannot be in the y function since the derivative of that involves the chain rule so it cannot be = - sqrt(t)y = -sqrt(t)*sqrt(t) e^( 2/3 t^(3/2))= -t e^( 2/3 t^(3/2))
The sqrt(t) needs to be taken out and then it works out just right.
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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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p(t) = 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?
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y= e^(-∫ 1/sqrt(t) dt) = e^(-2sqrt(t))
y'= -1/sqrt(t) e^(-2sqrt(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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-1/sqrt(t) e^(-2sqrt(t)) + 1/sqrt(t) e^(-2sqrt(t)) =0
This holds so we do have a solution.
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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'-1/t y = 0
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Why would we want to have done this?
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This puts our equation into the form we've been using and for which we have a method of solution.
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Imitating the reasoning we have seen, what is our y function?
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y= e^(ln t) = t
y'= 1
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Does it work?
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1- 1/t (t) = 1-1=0
So it does 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?
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Yes we've been integrating p(t) and then placing that function into the exponent.
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Will it always work?
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If we can put our equation into the form we have then yes.
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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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By the fundamental theorem of calculus:
y' = d(-int(p(t) dt)/dt e^(-int(p(t) dt)= -p(t) e^(-int(p(t) dt)
-p(t) e^(-int(p(t) dt)+ p(t) e^(-int(p(t) dt) = 0
This holds true.
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Is the equation satisfied?
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Yes, as demonstrated above.
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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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No, we have:
y'+sin(t) = 0
Which is not what we want, though easily solved.
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Is our equation therefore a homogeneous first-order linear equation?
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Since it doesn't fit it isn't homogenous.
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@& Equally importantly it's not first-order linear.*@
• t * y ' + t^2 y = 0
Once more, we need y ' to have coefficient t.
What is your conclusion?
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1/t(t * y ' + t^2 y = 0) = y'+ ty = 0
So yes, it's homogenous.
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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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Here:
y'= -tan(t) y
y' + tan(t)y = 0
So it's homogenous.
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@& More importantly it's first-order linear.*@
• y ' + t y^2 = 0
What do you think?
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No, because there's no way to reduce that y^2 term.
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• y ' + y = t
How about this one?
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Again, no. The right side cannot be combined with the y term to form p(t)y. This equation is first-order linear equation but it's not homogenous.
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@& right*@
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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The t in ty since p(t) = t in this case and ∫ t dt = 1/2 t^2
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• 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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By the chain rule:
d(e^(sin(t))*y)/dt = e^(sin(t))*y' + cos(t)*e^(sin(t))*y
Wait, multiplying by e^(sin(t)*y) won't work. There's an extra y in there that makes this incorrect. So it has to be e^(sin(t)).
(y'+ cos(t)*y)*e^(sin(t)) = y'*e^(sin(t))+cos(t)*y**e^(sin(t))
These are equivalent.
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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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We used the method of y= e^(-∫ p(t) dt) and ∫ p(t) dt = ∫ sin t dt= -cos 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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e^(t^2 / 2)* (y' + ty) = t* e^(t^2 / 2). Again we're using p(t) = t.
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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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This is because of the chain rule being used in differentiating the expression e^(t^2/2)*y wrt t.
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@& RIght, except that it's product rule, not chain rule.*@
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, C= constant.
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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=e^(t^2/2) + C
Dividing by e^(t^2/2) gives:
y= 1+ e^(-t^2/2)*C
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Is it a solution to the original equation?
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Yes, because:
y' = 0+ -t*e^(-t^2/2)*C
-t*e^(-t^2/2)*C + t(1+ e^(-t^2/2)*C) = -t*e^(-t^2/2)*C + t + t*e^(-t^2/2)*C)= t
This holds since the first and third terms on the left cancel each other.
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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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I think it has to be multiplied by e^(-integral(p(t) dt) not just the t integral of -p(t).
[y' + p(t)y = g(t)]*e^(-int(p(t)dt)) --> e^(-int(p(t)dt))*y' + p(t)*y*e^(-int(p(t)dt))= g(t)*e^(-int(p(t)dt))
We can see this is just what we had above where the left side is the result of the chain rule wrt t.
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*#&!
&#Your work looks good. See my notes. Let me know if you have any questions. &#