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course M 277
1/26 3Had trouble towards the end. Plan to go back and read the book and put more time in. Just wanted to have something submitted before class tomorrow.
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.
1)
Using chain rule for f(t) = e^(cos(t)), with y=f(t) and y=f(t)
f(t) = e^(g(t)) * g(t), where g(t) = cos(t) and g(t) = -sin(t)
=e^(cos(t))* -sin(t)
=-sin(t)e^(cos(t))
Plugging into equation y ' + sin(t) y = 0 we get,
-sin(t) e^(cos(t)) + sin(t) e^cos(t) = 0, which checks out
2)
Using chain rule for f(t) = e^(-t^3 / 3), with y=f(t) and y=f(t)
f(t) = e^(g(t)) * g(t), where g(t) = -t^3/3 and g(t) = -3*(t^2/3) = -t^2
=e^(-t^3/3)* -t^2
=-t^2 e^(-t^3 / 3)
Plugging into equation y ' + t^2 y = 0 we get,
-t^2 e^(-t^3 / 3) + t^2 e^(-t^3 / 3) = 0, which checks out
!!!! I just realized there was no need to show solution for 1st problem, only #2!!!!!!!!!!!!!!!!
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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.
For these questions we note that a coefficient of p(t) is attached to the variable y. We note that for fn e(p(t)) when derivative is taken we get fn p(t) which is which is coefficient of e base term and is equal but opposite of p(t). Pretty much meaning we need fn p(t) where when derivative is taken we get fn that is equal but opposite of p(t). Just by using fn of the form e^(p(t)), -p(t)) becomes coefficient for e. Which allows us to solve for homogeneous equation.
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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?
For y = sqrt(t) e^( 2/3 t^(3/2) )
y = (sqrt(t))* e^( 2/3 t^(3/2) ) + sqrt(t) *(e^( 2/3 t^(3/2) ))
sqrt(t) =t^(1/2) = (½)*(1/t^(1/2))
e^( 2/3 t^(3/2) ) = using chain rule,
e^(g(t))*g(t) = e^(2/3 t^(3/2))*t^(1/2), so that
y = (1/2*`sqrt(t))* e^( 2/3 t^(3/2) ) + sqrt(t) *(e^( 2/3 t^(3/2) )*`sqrt(t))
= [e^( 2/3 t^(3/2)/2`sqrt(t)] + [`sqrt(t)^2* e^( 2/3 t^(3/2)], factor out e^(2/3 t^(3/2))
= e^( 2/3 t^(3/2))[1/2`sqrt(t)+t]
??????Im not sure what Im missing, did you mean to put y = e^( 2/3 t^(3/2) and not y = sqrt(t) e^( 2/3 t^(3/2) )??????
If so
For y = e^( 2/3 t^(3/2),
y = `sqrt(t)*e^( 2/3 t^(3/2), without working it out because I already have a few lines back.
Pluggin into y ' - sqrt(t) y = 0, we get..
`sqrt(t) e^( 2/3 t^(3/2) - `sqrt(t) e^( 2/3 t^(3/2) = 0, which checks out.
????Im not sure if Im just not seeing what you want us to or if there has been a miss print but this makes sense to me. y = `sqrt(t)*e^(2/3*t^(3/2)) does not?????????????????????????????????????
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@& You are right. Should have been e^(2/3 t^(3/2) ).*@
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?
Coefficient of y was integrated in last problem, but note we were subtracting in last problem and adding y` and y in this one. So we have to take neg integral for solution to work, which is
.
-Int(1/`sqrt(t) dt) = Int( t^(-1/2) dt) = 2t^(1/2) or -2`sqrt(t)
Now for y = e^(-2`sqrt(t))
!!!!!!Im going to stop typing out every step for simple derivatives!!!!!!!!!!
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It's certainly OK to skip obvious steps. No point in wearing out your fingers.
If I need more detail than you're providing, I'll let you know.*@
y = -e^(-2`sqrt(t))/`sqrt(t) or (1/`sqrt(t))*e^(-2`sqrt(t)), when plugged into equation
y ' + 1/sqrt(t) * y = 0
[(1/`sqrt(t))*e^(-2`sqrt(t))] + [1/`sqrt(t))*e^(-2`sqrt(t))], which checks out
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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?
I answered this question in previous text area, sorry for inconvenience.
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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?
Again I already answered but I noted we had to account for Integral result being negative so terms would cancel out.
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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 (1/t)*y = 0
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Why would we want to have done this?
Isolate y from t and set equation equal tom 0 or have equation in homogeneous form.
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Imitating the reasoning we have seen, what is our y function?
Int( 1/t dt ) = ln(x)
Therefore y = e^(ln(x)),
y using chain rule
e^(ln(x))*ln(x) = e^(ln(x))*1/t
y=1/t*e^(ln(t))
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Does it work?
Plugging into equation y (1/t)*y = 0
(1/t)*e^(ln(t)) (1/t)*e^(ln(t)) = 0, which checks out.
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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?
Yes this exact method we have been using.
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Will it always work?
So far it has, but as long as you can find integral and equation is of form y + p(t)*y = 0 then it seems as though it should work.
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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?
y + p(t)e^(-Int(pt dt))
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Is the equation satisfied?
For y = e^(-int(p(t) dt)
y = e^(-int(p(t) dt)* (d/dt (-Int(p(t) dt)) = e^(-int(p(t) dt)*-p(t) = -p(t) e^(-int(p(t) dt), when plugged into equation
-p(t) e^(-int(p(t) dt) + p(t) e^(-int(p(t) dt) = 0, which checks out
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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?
Nope of the form y + sin(t) = 0, no y variable in 2nd term on left side.
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Is our equation therefore a homogeneous first-order linear equation?
No
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t * y ' + t^2 y = 0
Once more, we need y ' to have coefficient t.
What is your conclusion?
y + t*y = 0/t, where t is absorbed by 0. Which is homogeneous first-order linear 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.
y = -(sin(t)/cos(t))y
y+tan(t)*y = 0
Yes this is homogeneous first-order linear equation.
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y ' + t y^2 = 0
What do you think?
Yes this is homogeneous first-order linear equation.
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y ' + y = t
How about this one?
No this is homogeneous first-order linear equation because it does not equal 0.
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Solve the problems 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?
Int( t dt) = t^2/2, where t is coefficient of y.
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If you multiply both sides of the equation 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).
d/dt = (e^(sin(t) ))*y + e^(sin(t) )*y = cos(t)*e^(sin(t))*1 + e^(sin(t) )*0 = cos(t)*e^(sin(t))
?????Im confused are you talking about the equation cos(t) y ' = - sin(t) y, if so wouldnt our setup be???
???? y = (-sin*y)/cos = -sin/cos*y = -tan*y ????????????????????????????????????????????????
??????? y+tan*y = 0 ????????????????????????????
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@& y is an unknown function of t. Unless y is a constant function, y ' is not 0.
The term e^(sin(t) ) I y ' needs to remain in there.*@
How did we get e^(sin(t)) out of the expression y ' + cos(t)? Where did that sin(t) come from?
??????????????Im lost and not sure where I went wrong??????????????
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@& sin(t) is an antiderivative of cos(t).*@
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 do we get t^2 / 2 from the expression y ' + t y = t in the first place?
I get where t^2/2 comes from its the integral of the p(t) term, but not really understanding whats going on. I was understanding perfectly up til right now.
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@& The equation is y ' + t y = t .
Multiply both sides by e^(t^2 / 2).
Check that the left-hand side is indeed the derivative of e^(t^2 / 2) * y.
So when you integrate the left-hand side you get e^(t^2 / 2) * y.
The right-hand side is something you can integrate (and you do need to include the integration constant).
See what you get and send me a copy of the problem, with this note, and your inserted solution (and/or additional questions).*@
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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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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Set the results of the two integrations equal and solve for y. What is your result?
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Is it a solution to the original equation?
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If you multiply both sides of the equation y ' + p(t) y = g(t) by 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 do this.
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@& Check out my notes and submit that question, as well as any others you find you can do as a result. Or ask more questions, as indicated in the note.*@
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