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course MTH 279
6/14/13 9:54pm
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 = e^ (-t^3 / 3) then y' = -t^2 e^(-t^3 / 3)
y' + t^2 y = 0 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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They are all homogenous equations, in that they only have a single function of t, within them. This function is p(t). It can also be seen from y' + p(t)y = 0, that inorder to solve the equation we need y = e^ (-p(t)) and and that -p(t) must be equal to the antiderivative of p(t) given in the original solution
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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 a solution can not be found with that substition given the conditions, instead use y = e^ (2/3 t^(3/2)) the difference between the two is there is no sqrt(t) in the function
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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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sqrt(t) in the previous problem, in this problem it would be 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^ (-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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-1/sqrt(t) * e^ (-2 * sqrt(t)) + 1/sqrt(t) * e^ (-2 * sqrt(t)) = 0, yes a solution can be found, y' = -1/sqrt(t) * e^ (-2 * sqrt(t))
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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 gets the given problem into a common form that can be solved using known methods
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Imitating the reasoning we have seen, what is our y function?
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y = -1/t
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Does it work?
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yes
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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
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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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y' + p(t) e^(-int(p(t) dt) = 0
y' = -p(t) e^(-int(p(t) dt) substitute into previous function
-p(t) e^(-int(p(t) dt) + p(t) e^(-int(p(t) dt)
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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, it is in the form, y' + sin(t) = 0
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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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Divide both sides by t, we get y' + t y = 0 therefor it is a first in correct form and is a first-order homogeneous 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.
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y' + tan(t) y = 0, yes it is a first-order homogeneous equation
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y ' + t y^2 = 0
What do you think?
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No
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y ' + y = t
How about this one?
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No, this is a non-homogeneous
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Your 'yes' and 'no' answers are correct and I'm confident from your other answer that you know why, but you do need to at least briefly justify your choices.
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Solve the equations above that are homogeneous first-order linear equations.
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t * y ' + t^2 y = 0
y' + t y = 0
y = e^ (-int(t dt)
= e^ (- t^2/ 2) substitute back into the original
y' + t e^ (- t^2/ 2) = 0 y' can be found by subtracting the right hand side of the equation from both sides
-t e^ (- t^2/ 2) + t e^(- t^2/ 2) = 0
cos(t) y ' = - sin(t) y all steps are exactly like the ones above
y' + tan(t) y = 0
y = e^ (- int(tan(t) dt)
= e^ (- sec^2(x))
y' + tan(t) e^ (- sec^2(x)) = 0
-tan(t) e^ (- sec^2(x)) + tan(t) e^ (- sec^2(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 ' multiplier on ' 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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e^(sin(t)) y' + cos(t) e^(sin(t)) * y
(e^(sin(t)) * y)'
= (e^(sin(t))' y + e^(sin(t)) * y'
= cos(t) e^(sin(t)) * y + e^(sin(t)) *y'
e^(sin(t)) y' + cos(t) e^(sin(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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cos(t) which is p(t), using the equation e^(-int(p(t) dt)
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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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e^(t^2 / 2) y ' + t e^(t^2 / 2) y = t e^(t^2 / 2)
we get t^2 / 2 from t that multiplies the y, using the eqaution e^(- int(p(t) dt)
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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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well if we use, e^(t^2 / 2) * y, to get the left hand side which is just the derivative. It is mathematically possible to use the left hand side and integrate to return back to e^(t^2 / 2) * y, making it an anti-derivative
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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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yes it is
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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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e^(int(p(t) dt) y' + p(t) e^(int(p(t) dt) y = g(t) e^(int(p(t) dt)) shouldn't the integral have a negative sign in front of it???
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You need the negative sign for a first-order homogeneus.
For the integrating factor you don't need the negative sign.
If you think about this for a minute you'll see why. Having seen why you'll never need to remember which way it is, you'll be able to figure it out on the fly.
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e^(int(p(t) dt) y = int(g(t) e^(int(p(t) dt) dt)
take integral of both sides and then solve for y
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You appear to be in excellent shape here.
Do, however, check my inserted notes. Let me know if you have questions.
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