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course mth 279
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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If y = e^(-t^3 / 3) then y ' = -t^2 e^(-t^3/3) so that
y ' + t y becomes -t^2 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.
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All of the solutions use the exponential function because of the aforementioned trait. They are also all linear, homogeneous equations, which is why we can solve them in this fashion.
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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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Yes! If y = e^(-2/3t^(3 / 2)) then y ' = -t^(1/2) e^(-2/3t^(3/2)) so that
y ' + t y becomes -t^(1/2) e^(-2/3t^(3/2)) + t^(1/2) e^(-2/3t^(3/2)) = 0
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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 function, from the general form, p(t) to obtain the antiderivative P(t), in this case it is 1/sqrt(t) whereas before it was just p(t). Integrating p(t) in this case from c to t where P(t) is the particular antiderivative of p(t) that vanishes at the point t = c, we get P(t) = 2t^(-1/2)
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The derivative of 2 * t^(-1/2) is 2 * (-1/2) t^(-3/2), not 1 / sqrt(t).
Antiderivative of t^(-1/2) is 2 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^(-P(t)) = e^(-2t^(-1/2))
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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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If y = e^(-2t^(-1/2)) then y ' = -t^(-1/2) e^(-2t^(-1/2)) so that
y ' + t y becomes -t^(-1/2) e^(-2t^(-1/2)) + t^(-1/2) e^(-2t^(-1/2)) = 0
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Right idea, but you're making some errors in differentiation.
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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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First step: y = y/t
y - (1/t)*y = 0
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Why would we want to have done this?
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So that we can take a form that isnt standard for us and put the equation into a standard, clear, linear, homogeneous form.
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Imitating the reasoning we have seen, what is our y function?
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Our y function here would be logarithmic, particularly the natural log, found by taking the antiderivative of p(t) to yield P(t) = ln|t|, thus y = e^-lnt = t^(-1)
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Does it work?
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Yes, if y = t^(-1), y = -t^(-2), so that the equation y+ y/t = -t^(-2) + 1/t*1/t = 0
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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, this is what we have been doing. Finding P(t) from p(t) using integration and using this as our value for what we raise e to the power to in our equation for y, and thus in our equation for y
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Will it always work?
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No, it will not always work because it requires linear, homogeneous equations.
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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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- p(t) e^(-int(p(t) dt) + p(t) e^(-int(p(t) dt) = 0?
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Is the equation satisfied?
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It appears so! Both terms cancel each other and we are left with zero.
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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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Dividing both sides by y, we end up with y + sin(t) * (0 * y) = 0, y does not have a coefficient of one here. So our equation is not in the form y ' + p(t) y = 0, i.e., standard form.
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Is our equation therefore a homogeneous first-order linear equation?
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No it is not. It is homogeneous because of the 0 on the right side of the equation, but it is not linear because we do not have a linear value of y.
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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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This is fine, after dividing both sides by t, we end up with the following equation: y ' + t*y = 0 , which works because we have the zero on the right hand side, that is in the form y + p(t)y = g(t), g(t) = 0. And it is also linear because we have a coefficient of 1 on y.
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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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cos(t) y ' = - sin(t) y
Step one divide by cosine:
y ' = - sin(t) / cost(t) y
Step to move y over to the left side:
y ' + tan(t) y = 0,
This is a linear, homogeneous equation, p(t) = tan(t), and g(t) = 0
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y ' + t y^2 = 0
What do you think?
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g(t) = 0, but this is clearly non-linear because of the y^2 term.
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y ' + y = t
How about this one?
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Nope, we subtract t from both sides to get: y ' + y - t = 0, this is non-standard, linear form.
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Solve the equations above that are homogeneous first-order linear equations.
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y ' + t*y = 0
P(t) = 1/2t^2, If y = e^(-1/2t^2), y = -t e^(-1/2t^2). Putting the pieces together we obtain: - t e^(-1/2t^2) + t e^(-1/2t^2) = 0. Voila!
y ' + tan(t) y = 0
P(t) = -ln|cos(t)|, y = e^(ln|cos(t)|) = cos(t), y = -sin(t), so the equation becomes -sin(t) + tan(t)*cos(t) = -sin(t) + sin(t)*(cos(t)/cos(t)) = -sin(t) + sin(t) = 0. QED.
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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 p(t) portion of y+p(t)y = g(t). The t. We took the integral of it to get t^2/2
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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
Then multiplying both sides of the expression y ' + cos(t) y by e^(sin(t), we end up with = cos(t) e^(sin(t) * y + e^(sin(t) *y
The same result as the first part.
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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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Good question. It is the antiderivative of cos(t), or p(t) in the general form of the linear homogeneous equation.
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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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Ill start with the second question: t^2/2 came from the t in y + ty = t
Although on second thought Im not sure what to do with the t on the right side of the equation... stumped for now.
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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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I would say that this has to do with the product rule, we take the derivative of the first portion times the second portion, then add the derivative of the second portion to the first portion: e^(t^2 / 2) * y ' + t e^(t^2 / 2) y, or in general (f*g) = f*g + g*f
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This is the fundamental theorem of calculus. Any expression is an antiderivative of its derivative.
So if the left-hand side is the derivative of e^(t^2 / 2) * y, it follows that an antiderivative of the left-hand side is e^(t^2 / 2) * y.
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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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The integral is equal to e^(t^2 / 2), we could have used the u and v trick, but it was easy enough to see. And Wolfgram alpha confirmed this.
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Set the results of the two integrations equal and solve for y. What is your result?
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Sure thing:
e^(t^2 / 2) * y ' + t e^(t^2 / 2) y = e^(t^2 / 2)
Divide by e^(t^2 / 2):
y + ty = 1
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An antiderivative of the left-hand side is e^(t^2 / 2) * y.
So your integration would yield
e^(t^2 / 2) * y = e^(t^2 / 2) + c,
where c is an arbitrary integration constant.
What do you get if you solve this equation for y?
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Is it a solution to the original equation?
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I do not believe so, because the equation is non-homogeneous
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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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That is to say that y = e^(t^2/2), so (e^(t^2 / 2) * y) = e^(t^2 / 2) * y ' + t e^(t^2 / 2) y,
Multiplying both sides of y ' + p(t) y = g(t) by e^(t^2 / 2) gives us the same answer: e^(t^2 / 2) * y ' + t e^(t^2 / 2) y
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You're doing pretty well here.
I've inserted some comments.
This exercise should prepare you well for understanding the book's explanations and solving the problems in the problem set.
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I'll be glad to answer questions if you have them.
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