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course MTH 279
8/17 6
We show the following:y ' + t y = 0 has solution y = e^(-t^2 / 2).
If y = e^(-t^2 / 2) then y ' = -t e^(-t^2 / 2) so that
y ' + t y becomes -t e^(-t^2 / 2) + t e^(-t^2 / 2), 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^2e^(-t^3/3) so that
y’ + t^2y becomes -t^2e^(-t^3/3) + t^2e^(-t^3/3)
Confidence: 3
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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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The connection is that the derivative of e raised to a variable is the derivative of the variable times the original function. Once this derivative has been accomplished, you simply plug the new value into the given form, y’ + p(t)y = 0.
Confidence: 2
Self Critique:
I was slightly confused as to what connection the question is asking for. If I was correct in my thinking, I believe I explained it well.
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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 because it come out to:
sqrt(t)e^(2/3t^(3/2)) - te^(2/3t^(3/2)) = 0, which is not true.
Removing the sqrt(t) from the given y function would cause the formula to = 0.
Confidence: 2
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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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Following the previous process:
y’ = 1/sqrt(t) * y = 0
dy/dt = (-1/sqrt(t))y
integral(1/y)dy = integral(-1/sqrt(t))dt, this is the first thing we integrate.
Confidence: 2
Critique: The question was worded a bit confusing but I believe I answered correctly.
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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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Integral(1/y)dy = integral(-1/sqrt(t))dt, rewrite second part as -t^(-1/2)
ln(y) = -2t^(1/2)
e^(ln(y)) = e^(-2t^(1/2)), y = e^(-2t^(1/2))
Confidence: 3
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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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y’ + (1/sqrt(t))y = 0
y = e^(-2t^(1/2)), y’ = -t^(-1/2)e^(-2t^(1/2))
(-1/sqrt(t))e^(-2t^(1/2)) + (1/sqrt(t))e^(-2t^(1/2)) = 0
The statement is true so we do get a solution.
Confidence: 3
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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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ty’ = y
y’ = y/t, Divide both sides by t
y’ - y/t = 0, subtract right-hand side from both sides.
Confidence: 3
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Why would we want to have done this?
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It doesn’t appear to be necessary, however having y’ by itself would make the equation easier to work out. It allows you to have 0 on one side, making it simpler to recognize a true equation.
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Mostly, because this puts the equation into the same form as the others. The form is
y ' + (some function of t) * y = 0
In this case the 'some function of t' is - 1/t.
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Confidence: 2
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Imitating the reasoning we have seen, what is our y function?
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Although the order was changed, the y function remains the same, y = e^(-2t^(1/2))
Confidence: 1
Critique: I must be missing something here. I don’t really get what the question is asking, because the way I thought of this required no work as it was the same answer from two questions ago.
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I'm not sure where that came from either.
However you are getting this, so we'll move on.
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Does it work?
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(-1/sqrt(t))e^(-2t^(1/2)) - (1/t)(e^(-2t^(1/2))) = 0
No, however it would if it were instead y’ - y/sqrt(t).
Confidence: 3
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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, y’ is written as dy/dt and is then used to integrate the y and the p(t).
Confidence: 3
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Will it always work?
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Not necessarily, there may be extra unknown variables that cannot be integrated with current methods.
Confidence: 2
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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’ = e^(-int(p(t)dt)(-p(t))
Plugging into the equation:
-p(t)e^(-int(p(t)dt) + p(t)e^(-int(p(t)dt) = 0, Which is a true statement
Confidence:2
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Is the equation satisfied?
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Yes, it is a true statement.
Confidence: 2
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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. Having done this, the y in y’ + p(t)y = 0 is eliminated and we are left with y’ +p(t) = 0, with no dependence on y.
Confidence: 2
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Is our equation therefore a homogeneous first-order linear equation?
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No, it is still homogeneous first-order, but I believe it is non-linear.
Confidence: 2
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A linear equation needs both a y ' and a p(t) * y term, the sum being zero when the equation is homogeneous.
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The method used here works as long as we can integrate p(t).
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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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By dividing both sides by t, we are left with y’ + ty = 0, which is a homogeneous first order linear equation.
Confidence: 3
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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
y’ = -(sin(t)/cos(t))y
y’ + tan(t)y = 0, which is a homogeneous first-order linear equation.
Confidence: 3
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y ' + t y^2 = 0
What do you think?
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This equation is non-linear because of the y^2 term.
Confidence: 3
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y ' + y = t
How about this one?
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This equation is non-linear due to the placement of the t term.
Confidence: 2
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Solve the equations above that are homogeneous first-order linear equations.
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1. ty’ + t^2y = 0
y’ + ty = 0, rearranged form
dy/dt = -ty, rewrite the y’
(1/y)dy = -tdt, rearrange again
ln(y) = -(1/2)t^2 + c, make both sides a power of e, which eliminates the natural logarithm
y = Ae^(-(1/2)t^2), A representing an arbitrary constant
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Good solution, though you didn't solve it as a homogeneous first-order equation.
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2. cos(t)y’ = -sin(t)y, Showing slightly less steps than the last problem:
y’ + tan(t)y = 0, rearranged form
dy/dt = -tan(t)y, rewrite y’
ln(y) = ln(cos(t)) + c, rearrange and take the integral of each side, which eliminates nat. log.
y = Acos(t), final solution
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Again you didn't solve as a homogeneous first-order equation. Be sure you see how to do so, and check that your solutions are consistent with one another.
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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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t^2/2 is the antiderivative of p(t), which in this case is t.
Confidence: 3
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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)’ = e^sin(t)y’ + cos(t)e^sin(t)y
e^sin(t)(y’ + cos(t)y) = e^sin(t)y’ + cos(t)e^sin(t)y
Both expressions are equal.
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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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As learned earlier from the notes, when the expression is of form y’ + p(t)y, which it is here, we use e^(integral(p(t)dt)). In y’ + cos(t)y, cos(t) is p(t), and sin(t) is the antiderivative of cos(t).
Confidence: 3
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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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y’ + ty = t, multiply both sides by e^(t^2/2)
e^(t^2/2)y’ + e^(t^2/2)ty = te^(t^2/2), simplify left side
(e^(t^2/2)y)’ = te^(t^2/2), now integrating…
e^(t^2/2)y = e^(t^2/2) + c
t^2/2 came from the fact that we use the integral of p(t), which in this case is t.
Confidence: 3
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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 was shown in the first video on non-homogeneous first order linear equations. Derivatives and Antiderivatives are opposites of each other, so if you write something as say (xy)’, the antiderivative just eliminates the prime, leaving (xy).
Confidence: 3
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Right. In other words the original expression is an antiderivative of its 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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Integrating te^(t^2/2) gives e^(t^2/2) + c.
Confidence: 3
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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, divide out by e^(t^2/2)
y = 1 + c, here I am not sure if the right side should be 1 + ce^(-t^2/2), but I believe it is just c because it is an arbitrary constant and multiplying it by something would not change it.
Confidence: 2
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c is an arbitrary constant, but it still must be divided by e^(t^2 / 2).
So, as you can verify,
y = 1 + ce^(-t^2/2)
is the general solution to the equation.
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Is it a solution to the original equation?
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Yes, I believe it is the general solution.
Confidence: 2
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You can check by substituting the solution function into the equation.
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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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y’ + p(t)y = g(t), multiply both sides by e raised to the t integral of p(t)
e^(integral(p(t)dt))y’ + e^(integral(p(t)dt))p(t)y = e^(integral(p(t)dt))g(t), simplify left side
(e^(integral(p(t)dt))y)’ = e^(integral(p(t)dt))g(t)
Confidence: 3
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*#&!*#&!
&#Good work. See my notes and let me know if you have questions. &#