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course Mth 279
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.
y’ = -t^2(e^-t^3/3)
y’ + t^2 y becomes -t^2(e^(-t^3/3)) + t^2(e^(-t^3 / 3)) = 0
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.
The connection between these is that the derivative is equal, but opposite sign, to what is it being added/multiplied to.
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?
Yes this will give us a solution.
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?
1/sqrt(t) to get 2*sqrt(t)
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?
y = e^(2*sqrt(t))
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?
We get 0 if we plug this y into the equation.
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We don't get 0.
If we plug in y = e^(-2 sqrt(t)) we do get 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?
y’ - y/t = 0
Why would we want to have done this?
So we can get an equation like the last few examples we have done.
Imitating the reasoning we have seen, what is our y function?
y = e^(ln(t)) = t
Does it work?
Yes this equation works
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?
Yes it does.
Will it always work?
Yes it will always work.
What do you get if you plug y = e^(-int(p(t) dt) into the equation y ' + p(t) y = 0?
Is the equation satisfied?
Yes this equation is satisfied.
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You do need to show that is works. In general you always need to show your reasoning, at least in outline.
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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?
Is our equation therefore a homogeneous first-order linear equation?
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You don't seem to have answered the remaining questions in this document.
You will probably do so when you complete qa_2. I'll look forward to seeing your work.
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t * y ' + t^2 y = 0
Once more, we need y ' to have coefficient 1.
What is your conclusion?
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.
y ' + t y^2 = 0
What do you think?
y ' + y = t
How about this one?
Solve the equations 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?
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).
How did we get e^(sin(t)) from of the expression y ' + cos(t) y? Where did that sin(t) come from?
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?
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.
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.
Set the results of the two integrations equal and solve for y. What is your result?
Is it a solution to the original equation?
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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Good work on some of the early problems in this document.
You will want to complete the remaining problems, which you should do by submitting q_a_2.
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