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course Mth 279
oct 5 11
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 ????Shouldn’t this be y’ = -e^(-t)
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It should, and it is. The derivative of e^(-t) is (-t) ' e^(-t) = -1 * e^(-t) = - e^(-t).
That is, the derivative of -t is -1, not -t.
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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 becomes -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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All of these solutions come from first order linear differential equations that are homogeneous. This means that y ' + p(t) y is equal to zero.
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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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We do not get a solution. However, we will get the right solution in we use y = e^(2/3 t^3/2). This will give us sqrt(t) e^(2/3 t^3/2) - sqrt(t) e^(2/3 t^3/2) which equals zero.
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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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We integrated the sqrt(t) or the expression in front of the y.
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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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The integral of 1/sqrt(t) is 2sqrt(t) so our y would be 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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So if we use y = e^(2sqrt(t)), y’ = 1/sqrt(t) e^(2sqrt(t)) and plugging into the equation we get 1/sqrt(t) e^(2sqrt(t)) - 1/sqrt(t) e^(2sqrt(t)) which is equal to zero.
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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 = 0
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Why would we want to have done this?
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Because it gives us a homogeneous equation we can find solutions for.
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Imitating the reasoning we have seen, what is our y function?
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The integral of 1/t = ln(t) so our y functions is = e^(ln(t) ) which is just t.
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Does it work?
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For y = t, y’= 1 so plugging in we get 1- (1/ t) (t) =0
1-1 = 0
So yes this works.
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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 it does.
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Will it always work?
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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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This would mean that y’ = -p(t) e^(-int(p(t) dt) + p(t) e^(-int(p(t) dt) which = 0.
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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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No because dividing everything by y gets rid of the y variable next to P(t).
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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 1.
What is your conclusion?
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y’+ ty = 0
it is homogenous
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More completely is it homogeneous and first-order linear.
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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
it is homogenous
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... and first-order linear
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• y ' + t y^2 = 0
What do you think?
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This equation is not homogeneous
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It's equal to zero so it's homogeneous, and it's first-order, but it isn't linear.
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• y ' + y = t
How about this one?
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No this is not homogeneous
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It is first-order linear, but not homogeneous.
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Solve the equations above that are homogeneous first-order linear equations.
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For y’ + ty = 0 the y will equal e^(-t^2/2) so y’ = -t e^(-t^2/2). Plugging into the equation we get -t e^(-t^2/2) + te^(-t^2/2) = 0
For y’ + tan(t)y = 0
The integral of tan is -ln(cosx) so the y will equal -cosx and y’ = sinx
Plugging into the equation the cosx’s cancel so we get sin(x) - sin(x) = 0.
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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?
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The t that is in front of the y.
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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).
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The derivative of e^(sin(t))*y is e^(sin(t))*y’ + cos(t) e^(sin(t))*y
Which is the same as if you multiplied e^(sin(t))*y by y’ + cos(t).
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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?
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We took the integral of cos(t) which is sin(t) and put it in the exponent.
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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?
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From the integral of t.
You get e^(t^2 / 2) * y ' + t e^(t^2 / 2) y = t e^(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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Because the exponential derivative is the same as the function
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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 = e^(t^2/2) + c
y = 1 + c/e^( t^2/2)
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Is it a solution to the original equation?
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I am not sure
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The equation was
y ' + t y = t
y = 1 + c e^(-t^2 / 2)
so
y ' = -c t e^(-t^2 / 2)
Plug these into the equation and see if you get an identity. (you will)
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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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Using a random equation
The integral of 2t is t^2 so we will multiply e^(t^2) by both sides of the equation.
y’ + 2t*y = 4t
e^(t^2)*y’ + 2t e^(t^2)*y = 4t e^(t^2)
The left hand side of the equation turns out to be the derivative with respect to t of e^(t^2) * y just like the theorem states.
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&#Your work looks good. See my notes. Let me know if you have any questions. &#