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Messages - Hyunmin Jung

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When p(t) is continuous on the interval containing initial t

but g(t) is not continuous on the I containing initial t, it violates theorem 2.4.1 but is unsure whether or

not non-unique solution exist for all t in I? and the case for when g(t) is continuous on the interval and p(t) is not.

sorry, corrected.

If you are given two solutions that satisfies the ODE.

Without wronskian given, using two solutions and by using reduction of order,

is there a way where you are able to find third solutions to the equation?

I could not really get a clear answer from office hour and the problem I am referring is textbook p.228 #28

MAT244 Math--Lectures / Reference to term test 1
« on: October 06, 2014, 08:10:19 AM »
Is 2013S term test 1 an only reference that we are given to expect what will be on term test #1?

or is there more that I am missing out?

Also, What exactly is the point of abel's theorem rather than using matrices, is this just quick way of

achieving wronskian without having to derive the solution? How does the constant term for abel's theorem change??

MAT244 Math--Lectures / Distinguish between methods
« on: October 03, 2014, 10:40:15 AM »
I can solve the problems but I have trouble distinguishing between methods. 2.1 #38. Why is this method called a variation of parameters? Is "Integrating Factor  where adding mu to the equation" a part of variation of parameters (Since you are technically adding a parameter to  p(t))? I always thought variation of parameters as adding some coefficient on complementary equation to find other fundamental solution of y to derive general solutions.

MAT244 Math--Lectures / 3.1 #2
« on: September 30, 2014, 02:32:14 AM »
Finding the general solution of the given differential equation
$$y" + 3y' + 2y = 0 $$
assuming $y = e^{rt}$ is a solution (I have not fully understand why $y = e^{rt}$ is a solution) to the derive characteristic equation.  Can I get some clear explanation?
We are looking the solution in this form (intuition)--V.I.
 I understand that it satisfies equation such as $y" - y = 0$) Nope unless $r=\pm 1$
y" = r^2e^{rt},\\
y' = re^{rt},\\
y  = e^{rt.}
Characteristic Equation $r^2+3r+2 = 0$.

Roots are $r_1=-1 and  $r_2=-2$
y_1 = e^{-t},\\
y_2 = e^{-2t},\\
y = c_2e^{-t}+c_2*e^{-2t}

MAT244 Math--Lectures / 1.2 8 a)
« on: September 15, 2014, 03:08:15 AM »
Can someone verify that this is correct?

dp/dt = rp

a) Find the rate constant r if the population doubles in 30 days.

ln lpl = rt + C
lpl = e^(rt+ C)
p = e^rt*+-e^C
Set +- e^C = +c
p = c*e^rt

initial condition at t = 0

p0 = c*e^0
c = p0

population doubles in 30 days
let t be number of days

2p0 = p0(e^30r)
2 = e^30r
ln(2) = 30r
r = ln(2)/30

MAT244 Math--Lectures / Graphing direction field in general
« on: September 15, 2014, 01:12:24 AM »
It's been a long time since I have taken past math courses so I am not really sure what's going on in some of the content.

My question is for 1.1.31

Okay, I drawn a direction field but I am not sure where I go from there.

What exactly does the solution mean?? Are the solution just slope of the y in general?

there is not a diverging point where vector changes direction (since in this question direction goes all over the place except when it is fixed sign when t is negative) so I guess it does not diverge in this question?

How can you tell where there is asymptote? Do you have set the slope to be 0 before deriving the asymptote?? If so, Why??
By factoring it does come up to be + or - root of (2t-1) But I am assuming because the sign is all negative where t < 0 so there is only one asymptote at + root of (2t-1)? Is my logic correct?

Thanks in advance

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