#$&*
course MTH 279
I was confused on how this one needed to be turned in, seemed as if you wanted this set of notes as an assignment per instructions on the assignment page I may be wrong.
The **** mark and the #$&* mark should each appear by itself, on its own line.We show the following:
• y ' + t y = 0 has solution y = e^(-t^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.
So that y ' + t^2 y becomes -t*e^(-t^3/3) + t^2*(e^(-t^3/3) = 0
#$&*
What do all three solutions have in common?
Some of this is left to you.
However for one thing, note that they all involve the fact that the derivative of a function of form e^(-p(t)) is equal to -p'(t) e^(-p(t)).
And all of these equations are of the form y ' + p(t) y = 0.
Now you are asked to explain the connection.
All of these solutions come from first order linear differential equations that are homogeneous. This means that y ' + p(t) y is equal to zero
#$&*
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?
We do not get a solution:
Substituting we get = (t) e^( 2/3 t^(3/2) ) - sqrt (t) * sqrt(t) e^( 2/3 t^(3/2) ) = 0
0=0
We could make it separable differential equation setting the y’s on one side and the t’s on the other. We will get the right solution if we use y = e^(2/3 t^3/2
#$&*
• sqrt(t) y ' + y = 0?
The rest of our equations started with y ' . This one starts with sqrt(t) y '.
We can make it like the others if we divide both sides by sqrt(t).
We get
• y ' + 1/sqrt(t) * y = 0.
Follow the process we used before.
We first integrated something. What was it we integrated?
Integrated one side with the y by dy and one with the t with respect to dt
#$&*
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?
The integral of 1/sqrt(t) is 2sqrt(t) so our y would be e^(2sqrt(t))
#$&*
What do we get if we plug our y function into the equation? Do we get a solution? If not, how can we modify our y function to obtain a solution?
If we use y = e^(2sqrt(t)), y’ = 1/sqrt(t) e^(2sqrt(t)) and plugging into the equation we get 1/sqrt(t) e^(2sqrt(t)) - 1/sqrt(t) e^(2sqrt(t)) which is equal to zero
#$&*
• 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
#$&*
Why would we want to have done this?
To put the equation in a more workable form based on what we have already done.
#$&*
Imitating the reasoning we have seen, what is our y function?
e^ (ln(t)) = t
#$&*
Does it work?
y = t, y’= 1 so plugging in we get 1- (1/ t) (t) =0
1-1 = 0
It works
#$&*
• 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, if I am understanding it correctly
#$&*
Will it always work?
yes
#$&*
What do you get if you plug y = e^(-int(p(t) dt) into the equation y ' + p(t) y = 0?
- int(e^(-int(p(t))dt)) + p(t) e^(-int(p(t) dt) = 0
#$&*
Is the equation satisfied?
Yes the equation is satisfied
#$&*
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?
No because if one divides by “y” then y is cancelled in the second part of the equation
#$&*
Is our equation therefore a homogeneous first-order linear equation?
No this is not
#$&*
• t * y ' + t^2 y = 0
Once more, we need y ' to have coefficient 1.
What is your conclusion?
Yes this is a first order linear homogenous equation
#$&*
• cos(t) y ' = - sin(t) y
Again you need y ' to have coefficient 1.
Then you need the right-hand side to be 0.
Put the equation into this form, then see what you think.
y’ + tan(t) y
it is a first order linear differential equation
#$&*
• y ' + t y^2 = 0
What do you think?
This is not linear, but it is a first order homogenous equation
#$&*
• y ' + y = t
How about this one?
This is is linear first order equation but not a homogenous equation because there is an additional -t on the left side
#$&*
Solve the equations above that are homogeneous first-order linear equations.
1. t * y ' + t^2 y = 0
y’ + ty = 0
y = e^(t^2/2)
y’ = t e^(t^2/2)
t e^(t^2/2) + t e^(t^2/2) = 0
2. cos(t) y ' = - sin(t) y
y’ + tan(t) y = 0
y = e^(-ln(cos(t)))
y’ = -tan ( e^ (-ln(cos(t))))
-tan ( e^ (-ln(cos(t)))) + tan(t) e^(-ln(cos(t)))
#$&*
Verify the following:
• If you multiply both sides of the equation y ' + t y by e^(t^2 / 2), the result is the derivative with respect to t of e^(t^2 / 2) * y.
The derivative with respect to t of e^(t^2 / 2) * y is easily found by the product rule to be
(e^(t^2 / 2) * y) '
= (e ^ (t^2 / 2) ) ' y + e^(t^2/2) * y '
= t e^(t^2/2) * y + e^(t^2 / 2) * y '.
If you multiply both sides of y ' + t y by e^(t^2 / 2) you get e^(t^2 / 2) y ' + t e^(t^2 / 2).
Same thing.
Now, what is it in the original expression y ' + t y that led us to come up with t^2 / 2 to put into that exponent?
y’ + ty
#$&*
• 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).
The derivative of e^(sin(t))*y is e^(sin(t))*y’ + cos(t) e^(sin(t))*y
Which is the same as if you multiplied e^(sin(t))*y by y’ + cos(t).
#$&*
How did we get e^(sin(t)) from of the expression y ' + cos(t) y? Where did that sin(t) come from?
When we took the integral of t we found it was equal to sin(t)
#$&*
• 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 did we get t^2 / 2 from the expression y ' + t y = t in the first place?
The answer is given below so I was uncertain if I needed to answer with the same answer.
#$&*
The equation becomes e^(t^2 / 2) * y ' + t e^(t^2 / 2) y = t e^(t^2 / 2).
The left-hand side, as we can easily see, is the derivative with respect to t of e^(t^2 / 2) * y.
So if we integrate the left-hand side with respect to t, since the left-hand side is the derivative of e^(t^2 / 2) * y, an antiderivative is e^(t^2 / 2) * y.
Explain why it's so.
It is so because if one were to integrate t, then one would raise the power divide by the new power and in a first order linear equation, it is of the form e^ (-int(t))
#$&*
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
#$&*
Set the results of the two integrations equal and solve for y. What is your result?
e^( t^2/2) *y = e^(t^2/2) + C divide each side to isolate y
y = 1 + C/e^( t^2/2)
#$&*
Is it a solution to the original equation?
y ' = -c t e^(-t^2 / 2)
so yes it is the identity
#$&*
• If you multiply both sides of the equation y ' + p(t) y = g(t) by the e raised to the t integral of p(t), the left-hand side becomes the derivative with respect to t of e^(integral(p(t) dt) ) * y.
See if you can prove this.
The integral of 2t is t^2 so we will multiply e^(t^2) by both sides of the equation
y’ + 2t*y = 4t
e^(t^2)*y’ + 2t e^(t^2)*y = 4t e^(t^2)
The left hand side of the equation turns out to be the derivative with respect to t of e^(t^2) * y
#$&*
"
&#This looks good. Let me know if you have any questions. &#