#$&*
course MTH 279
1/24 2:36PM
This document and the next are supplemented by Chapter 2 of the text.This should be submitted as a q_a_ document, filling in answers in the usual manner, between the marks
****
#$&*
The **** mark and the #$&* mark should each appear by itself, on its own line.
We show the following:
y ' + t y = 0 has solution y = e^(-t).
If y = e^(-t) then y ' = -t e^(-t) so that
y ' + t y becomes -t e^-t + t e^-t, which is zero.
y ' + sin(t) y = 0 has solution y = e^(cos t)
If y = e^(cos t) then y ' = -sin(t) e^(cos(t)) so that
y ' + t y becomes -sin(t) e^(cos(t)) + sin(t) e^cos(t) = 0
y ' + t^2 y = 0 has solution y = e^(-t^3 / 3)
This is left to you.
****
y= -t^2 e^(-t^3/3)
-t^2 e^(-t^3/3) + t^2 e^(-t^3 / 3) = 0
#$&*
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 relationship is that
y is always e^-int(p(t) dt)
#$&*
What would be a solution to each of the following:
y ' - sqrt(t) y = 0?
If we integrate sqrt(t) we get 2/3 t^(3/2).
The derivative of e^( 2/3 t^(3/2) ) is t^(1/2) e^ ( 2/3 t^(3/2) ), or sqrt(t) e^( 2/3 t^(3/2) ).
Now, if we substitute y = sqrt(t) e^( 2/3 t^(3/2) ) into the equation, do we get a solution? If not, how can we modify our y function to obtain a solution?
****
No we do not get a solution. Because our outcome is
y= sqrt(t) e^(2/3 t^3/2) so y = t + 1/2sqrt(t) e^(2/3 t^3/2)
giving us equation: t + 1/2sqrt(t) e^(2/3 t^3/2) - sqrt(t) (sqrt(t) e^(2/3 t^3/2)) = 0
which is not true. We would need to remove the sqrt(t) from y in order for this equation to work. Doing so would result in:
sqrt(t) e^(2/3 t^3/2) - sqrt(t) e^(2/3 t^3/2) = 0 ; which is true.
#$&*
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?
****
The only way I see this working is if 1/sqrt(t) was -1/sqrt(t). Here is what happens otherwise
Integrate 1/sqrt(t) = 2sqrt(t)
solve equation: (e^2sqrt(t))/sqrt(t) + 1/sqrt(t) e^2sqrt(t)
=> (e^2sqrt(t))/sqrt(t) +(e^2sqrt(t))/sqrt(t)= 0 which is false
If the + was - this we our function would be (e^2sqrt(t))/sqrt(t) - (e^2sqrt(t))/sqrt(t) = 0 which is true.
#$&*
@&
Your solution would be
y = e^(-2 sqrt(t)),
which you can verify gives you a solution.
*@
We then formed an exponential function, based on our integral. That was our y function. What y function do we get if we imitate the previous problem?
****
Exponentialize 2sqrt(t) = e^2sqrt(t)
Set = to y = e^2sqrt(t)
#$&*
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?
****
y= (e^2sqrt(t))/sqrt(t)
solve equation: (e^2sqrt(t))/sqrt(t) + 1/sqrt(t) e^2sqrt(t)
=> (e^2sqrt(t))/sqrt(t) +(e^2sqrt(t))/sqrt(t)= 0 which is false
If the + was - this we our function would be (e^2sqrt(t))/sqrt(t) - (e^2sqrt(t))/sqrt(t) = 0 which is true.
Therefor we would need to modify our y to be y = e^-2sqrt(t)
Where our y would be -e^-2sqrt(t)/ sqrt(t)
Which would change our function to be;
-e^2sqrt(t)/sqrt(t) + e^2sqrt(t)/sqrt(t)= 0 which his indeed true.
#$&*
t y ' = y?
If we divide both sides by t and subtract the right-hand side from both sides what equation do we get?
****
y - y/t = 0 or y - y * 1/t = 0
#$&*
Why would we want to have done this?
****
We want to do this because now the function is in the form weve been working with throughout this assignment.
#$&*
Imitating the reasoning we have seen, what is our y function?
****
Integral of 1/t = ln(|t|) now exponentialize and set equal to y
Y= e^ln(|t|) ; find y
Y = |t|
Solve; |t| - e^ln(|t|)/t = 0
#$&*
Does it work?
****
Yes because e^ln(|t|)/t = |t|
Therefore we have |t| - |t| = 0 which is true.
#$&*
y ' + p(t) y = 0 has solution y = e^(- int(p(t) dt)).
This says that you integrate the p(t) function and use it to form your solution y = e^(- int(p(t) dt)).
Does this encapsulate the method we have been using?
****
Yes it does.
#$&*
Will it always work?
****
It certainly seems that it does.
#$&*
@&
It will indeed work, as long as p(t) can be integrated.
*@
What do you get if you plug y = e^(-int(p(t) dt) into the equation y ' + p(t) y = 0?
****
-p(t) e^(-int(p(t) dt) + p(t) e^(-int(p(t) dt) = 0
#$&*
Is the equation satisfied?
****
Yes it is.
#$&*
y ' + p(t) y = 0 is the general form of what we call a first-order linear homogeneous equation. If it can be put into this form, then it is a first-order linear homogeneous equation.
Which of the following is a homogeneous first-order linear equation?
y * y ' + sin(t) y = 0
We need y ' to have coefficient 1. We get that if we divide both sides by y.
Having done this, is our equation in the form y ' + p(t) y = 0?
****
If you divide both sides by y our equation would be y + sin(t) = 0 which is not in form that we need.
#$&*
Is our equation therefore a homogeneous first-order linear equation?
****
I would have to say no.
#$&*
t * y ' + t^2 y = 0
Once more, we need y ' to have coefficient 1.
What is your conclusion?
****
Divide both sides by t = y + ty = 0 which is a homogeneous equation.
#$&*
@&
Not just homogeneous, but first-order linear homoegeneous.
*@
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)y)/cos(t)
y + (sin(t) y)/cos(t) = 0
equation is homogeneous
#$&*
@&
Again, first-order linear homoegeneous.
First-order because the highest derivative is the first.
Linear because y and y ' both appear as first powers.
Homogeneous because the right-hand side is zero.
*@
y ' + t y^2 = 0
What do you think?
****
Not homogeneous
#$&*
@&
It's homogeneous but it's not linear because y^2 is not a first power of y.
*@
y ' + y = t
How about this one?
****
Dont believe this is homogeneous either. Because if we subtract t from both sides we get y + y - t = 0 which would not be in the form that we need.
#$&*
@&
This is first-order and linear but not homogeneous.
*@
Solve the equations above that are homogeneous first-order linear equations.
****
y + ty = 0 -int (t) dt = t^2/2
y = e^(-t^2/2)
y = t e^(-t^2/2)
-te^(t^2/2) + t e^(t^2/2) = 0 which is true
And,
y + sin(t)/cos(t) * y = 0 -int sin(t)/cos(t) = ln(|cos(t)|)
y= e^ln(|cos(t)|)
y = -sin |cos(t)|
solution;
-sin |cos(t)| + sin(t)/cos(t) * e^ln(cos(t) = -sin |cos(t)| + sin |cos(t)| = 0
#$&*
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?
****
The integral of t led us to t^2/2
#$&*
If you multiply the expression y ' + cos(t) y by e^(sin(t) ), the result is the derivative with respect to t of e^(-sin(t)) * y.
Just do what it says. Find the t derivative of e^(sin(t) ) * y. Then multiply both sides of the expression y ' + cos(t) y by e^(sin(t) * y).
****
T derivative of e^(sin(t)) * y =
cos(t) e^(sin(t)) * y + e^(sin(t)) y
multiply both sides by e^(sin(t)) = e^(sin(t)) y + cos(t) e^sin(t) y
#$&*
How did we get e^(sin(t)) from of the expression y ' + cos(t) y? Where did that sin(t) come from?
****
e^(sin(t)) is the integral of cos(t) that has been exponentialized.
#$&*
If you multiply both sides of the equation y ' + t y = t by e^(t^2 / 2), the integral with respect to t of the left-hand side will be e^(t^2 / 2) * y.
You should have the pattern by now. What do you get, and how did we get t^2 / 2 from the expression y ' + t y = t in the first place?
****
We get t^2/2 by the integral of t
#$&*
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.
****
Since we acquired e^(t^2 / 2) * y ' + t e^(t^2 / 2) y by differentiating e^(t^2/2) * y then we can assume that the integral of e^(t^2 / 2) * y ' + t e^(t^2 / 2) y would be
e^(t^2/2) * y
#$&*
Having integrated the left-hand side, we integrate the right-hand side t e^(t^2 / 2).
What do you get? Be sure to include an integration constant.
****
e^(t^2/2) + c
#$&*
Set the results of the two integrations equal and solve for y. What is your result?
****
e^(t^2/2) * y = e^(t^2/2) + c ; divide both sides by e^(t^2/2)
y= (e^(t^2/2) + C)/e^(t^2/2)
#$&*
Is it a solution to the original equation?
****
Yes it is a solution to the problem.
y= (e^(t^2/2) + C)/e^(t^2/2)
therefore; y = 0
t = e^(t^2/2) so our equation is;
0 + e^(t^2/2) * (e^(t^2/2) + C)/e^(t^2/2) = e^(t^2/2)
e^(t^2/2) + C = e^(t^2/2) which works if C = 0
#$&*
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.
****
Int(p(t) dt) = p(t)^2/2 then; exponentialize
e^(p(t)^2/2) find derivative of e^(p(t)^2/2) * y
e^(p(t)^2/2) * y + p(t) e^(p(t)^2/2) * y
if we multiply both sides of y + p(t) y = g(t) by e^(p(t)^2/2) the left had side of the equation will be equal to the derivative of e^(p(t)^2/2) * y
so;
e^(p(t)^2/2)(y + p(t) y) = g(t)
e^(p(t)^2/2) * y + p(t)[e^(p(t)^2/2)]*y = g(t) e^(p(t)^2/2)
We see that the left hand side is indeed the same as the derivative of e^(p(t)^2/2) * y
#$&*
@&
Very good. Check my notes to clarify some terminology.
*@