Homework 8 collaboration area
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From Mnestero:
So after a bunch of algebra to solve the system of equations on prob 12 of 6.7 I got an answer. I often make simple mistakes, so I wanted to see if anyone else got what I have:
y1 = cos(sqrt(2)t)+ 2/5 cos(t)- 7/5 cos(sqrt(6)t) y2 = 1/5 cos(t) + 14/5 cos(sqrt(6)t)
From Eun Young:
If you hit the system of differential equations by the Laplace transform, you'll get
S2Y1 − S = − 2Y1 + 2Y2 and S2Y2 − 3S = 2Y1 − 5Y2. This is a system of two equations. Solve this for Y1 and Y2 using Cramer's rule or just algebra. Then,
$ Y_1 = \frac{s^3+11s}{s^4+7s^2+6} $. Find y1 using partial fractions. It's similar for y2
I'm trying to complete the partial fractions for Y1 and am getting stuck at "B = -sqrt(6) i (1-A)." Any thoughts? I'm not able to break down the values of A and B (or C and D) unless there is another step I've missed. Thanks. -Christine
Suggestion from Steve Bell: Christine, you shouldn't be getting complex numbers there. Notice that the denominator factors
(s2 + 1)(s2 + 6).
So your partial fractions terms should look like this:
$ \frac{As+B}{s^2+1}+\frac{Cs+D}{s^2+6} $
where A,B,C,D will be real valued.
Response from Mickey Rhoades Mrhoade
Been on vacation and finally gettin gback on the Rhea. For problem 12, I solved the partial fractions the way Dr. Bell suggested: A=6/5, B=4, C=-6/5, D=1. y(t) = 3/5 cos(t) + 2 sin(t) +12/5 cos (sqrt(6)t) - sqrt (2/3) sin (sqrt(6) t)
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From Chris:
The example in the book and in our notes doesn't look the same as the problem 5 in 12.12. I'm not even sure how to set up the problem. Can anyone help get me started?
From Mnestero:
I started this problem by taking the laplace with respect to t. This gave me s*W = x/s^2 - x d/dx * W. I took the derivative of x to be 1. I then solved for W, which gave me W = x/(s^2(s+1)). After this use partial fractions. I am not positive that this is the correct approach - but it matches the answer in the back of the book. Anyone else have any thoughts?
From Eun Young:
If you take the Laplace transform with respect to t you'll have
$ x \frac{\partial W}{\partial x} + s W = \frac{x}{s^2} $.
Divide both sides by x then you'll have
$ \frac{\partial W}{\partial x} +\frac{ s}{x} W = \frac{1}{s^2} $.
This is a first-order linear ODE. Be careful. s is a constant. See section 1.5 for reference.
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Here's a little refresher about how to solve first order linear ODEs. Steve Bell
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On prob 16 of 11.1, I set up three piece wise functions to find Bn (odd function An and Ao are 0). From -pi to -pi/2 I set F(x)=0 From -pi/2 to pi/2 I set F(x)=x From pi/2 to pi I set F(x)=0
From here I solved for Bn. The fourier series I calculated is F(x) = 2/pi sinx - 2/4pi sin2x + 2/9pi sin3x ... When graphing this, it is similar to the original graph, but seems slightly off. Am I setting up the problem wrong?
From Andrew:
I don't believe we have to do prob 16 of section 11.1 for the homework, only 11.1.12, 11.1.14, and 11.1.18
From Michael:
Yeah, you're right. Well I guess I got some extra practice in.
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I suppose there is a more elegant way to get through 6.6 #8 that direct integration...is this valid?
f(t) = sint
Use shifting theorem, so from the table, (1+k)/(s^2 + (1+k)^2)
and by differentiation of transforms, L{tf(t)} = - derivative( (1+k)/(s^2 + (1+k)^2) ) ?
Also wondering about #16. I have done a bit of algebra to get to the form (2s+6)/ (s^2 (s+6)^2 + 20 (s+5)(s+1))
My CAS tells me the answer is e^-3t t sin t but I don't see it in there.
Thanks!
-Christine
From Ryan Leemhuis:
Hi Christine,
The way I approached this problem was to recognize that we have a transform identity for e^-(kt)*sin(t) in the identity table of 6.1. Using the differentiation theorem of laplace transforms we know that multiplying the function by t acts as the negative differentiation of the laplace transform. All you should have to do is take the derivative of the identity given in the table and multiply by -1.
In regards to problem #16, the key for me was attempting to find similarity between the numerator and denominator. In my case, that was creating a (s+3) term in both. Once this is done, you may recognize the form from the previous problems done in this section. Consider what common form in the table can be differentiated to create this form. Keep in mind that when you differentiate a fraction like we have it often results in a squaring of the entire denominator.
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Question from Luo Shibo
for problem 12.12, I convert the problem to the ODE below;
$ \frac{\partial W}{\partial x} +\frac{ s}{x} W = \frac{1}{s^2} $.
I treat s as a constant, and x as variable.
Since there is $ \frac{ s}{x} W $,I find it's difficult for me to solve this ODE, can any one give me some help?
From Eun Young:
You have a first-order ODE in the form of
$ \frac{\partial W}{\partial x} + p(x) W = r(x) $ where $ p(x) = \frac{s}{x} , r(x) = \frac{1}{s^2} $.
You can solve this by multiplying both sides by an integrating factor $ F = e^{\int p(x) dx} = e^{ s \ln(x) } = (e^{ \ln(x)})^s = x^s. $ Then, you'll have $ F \frac{\partial W}{\partial x} + p F W = r F $. Note that $ pF = \frac{\partial F}{\partial x}. $
Hence, we have $ \frac{ \partial}{\partial x} (FW) = rF. $ If you integrate both sides with respect to x you will get FW and use initial or boundary conditions to find W.
See section 1.5. There is detailed explanation.
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From Ben Minchhtoff
The right side of 12.12 #5 is xt to start. Shouldn't the Laplace transform of that be x over s squared? If that is the case dividing thorugh by x should give you 1 over s squared on the right not x over s squared?
From Eun Young :
Yes, you're right. I corrected it. Thanks, Ben!
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From Hunter Zillmer
For 11.1, when it says plot the partial sums does it mean plot them as individual functions on the same plot or does it mean plot the function as the partial sums summed together?
Thanks!
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From Steve Bell:
Hunter, I think they want you to graph the sum of the Fourier series up to the n=5 terms. By the way, to plot multiple functions on a single graph, here's a great example in MAPLE:
plot( { x , x-x^3/6 , x-x^3/6+x^5/5! , sin(x) } , x=0..2*Pi);
This would give you a plot of those four functions on a single graph and would demonstrate that the power series for the sine function starts hugging the graph of sine right away.
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