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Differential Equations Help

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2 minutes ago, Dash Lambda said:

No, the real part governs the damping, and damping doesn't change the frequency of oscillation.

When the root is complex, the exponential becomes e(a + bi)t, which you separate into eate(bi)t. When you substitute using Euler's identity, you only replace the exponential containing the imaginary number, so only b influences the frequency.

I'm a bit confused. Would the frequency would just be .5508hz? Looking at the plot, it looks like the frequency should be more like 2.something~

 

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4 minutes ago, djdwosk97 said:

I'm a bit confused. Would the frequency would just be .5508hz? Looking at the plot, it looks like the frequency should be more like 2.something~

From the range displayed on that plot, the first complete cycle is ~4.281 to ~6.096 (in terms of t), which means it has a period of ~1.815.

This converts to a frequency of 1/1.815 =~ .551hz, or a circular frequency of 2pi/1.815 =~ 3.4618 rad/s which, accounting for roundoff error, agrees with the circular frequency of 3.46085 present in the solution.

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I might have have taken Diff Eq had the professor not been someone that can barely speak English. Ended up taking Finite Math instead and currently Calc.

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22 minutes ago, Dash Lambda said:

From the range displayed on that plot, the first complete cycle is ~4.281 to ~6.096 (in terms of t), which means it has a period of ~1.815.

This converts to a frequency of 1/1.815 =~ .551hz, or a circular frequency of 2pi/1.815 =~ 3.4618 rad/s which, accounting for roundoff error, agrees with the circular frequency of 3.46085 present in the solution.

Ahhhh, I thought period = frequency. Okay, got it. 

 

How would I calculate the amplitude then, or is it just one of the terms in the equation (yformula[t])? 

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28 minutes ago, djdwosk97 said:

How would I calculate the amplitude then, or is it just one of the terms in the equation (yformula[t])? 

Since sin() and cos() represent the side lengths of a right triangle, the amplitude of the undamped oscillation is the hypotenuse. You just plug the coefficients into the Pythagorean theorem.

And as this oscillation is damped, the actual amplitude will be the undamped amplitude times the damping function (since the damping just 'squeezes' the wave).

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6 minutes ago, Dash Lambda said:

Since sin() and cos() represent the side lengths of a right triangle, the amplitude of the undamped oscillation is the hypotenuse. You just plug the coefficients into the Pythagorean theorem.

And as this oscillation is damped, the actual amplitude will be the undamped amplitude times the damping function (since the damping just 'squeezes' the wave).

So what would the dampening function be? 5.7y'[t]? 

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6 minutes ago, djdwosk97 said:

So what would the dampening function be? 5.7y'[t]? 

It's the exponential part of the solution.

For an oscillation of the form eat(C1sin(bt) + C2cos(bt)), you have the simple oscillation C1sin(bt) + C2cos(bt) with a fixed amplitude, and the damping function eat that squeezes the wave.

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9 minutes ago, Dash Lambda said:

It's the exponential part of the solution.

For an oscillation of the form eat(C1sin(bt) + C2cos(bt)), you have the simple oscillation C1sin(bt) + C2cos(bt) with a fixed amplitude, and the damping function eat that squeezes the wave.

Perfect, thanks so much. 

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@Dash Lambda

Sorry to keep asking questions, but I'm taking an online diffeq course, so I don't really have a professor/TA to ask, and I don't seem to have much luck searching for the answers online, and the one textbook I found doesn't really help explain things (at least in a way that I understand them). And your answers do really help me understand what I'm doing.

------

 

I need to find the amplitude in a resonating unforced oscillating diffeq, f[t] = yunforced[t].

y[t] = (3.6-.1t)Cos[2t] + (.9+1.8t)Cos[t]Sin[t] 

 

So would the amplitude just be the square root of the coefficients squared/2? So: 1/2 * ((3.6-.1t)^2 + (.9+1.8t)^2)^.5

 

------

 

And the frequency of a resonating unforced oscillator would be the same as the frequency of the (non-resonating) unforced oscillator? because resonance only affects amplitude, right? 

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11 hours ago, DeadEyePsycho said:

I might have have taken Diff Eq had the professor not been someone that can barely speak English. Ended up taking Finite Math instead and currently Calc.

I had the same problem with both college algebra and some basic chemistry class.

 

I failed the math class and almost failed the chem class I'm ashamed to say.

 

Though I partly blame that on the German t.a. teaching the algebra class for having too much of an accent and on the worst math textbook I've ever seen.

 

As for the chem class I partly suck at chem and the t.a. for that was Indian (from India) and spoke with such a heavy accent I could barely tell he was speaking English.  He was way too quiet too so I had trouble hearing him.

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10 hours ago, djdwosk97 said:

I need to find the amplitude in a resonating unforced oscillating diffeq, f[t] = yunforced[t].

y[t] = (3.6-.1t)Cos[2t] + (.9+1.8t)Cos[t]Sin[t] 

 

So would the amplitude just be the square root of the coefficients squared/2? So: 1/2 * ((3.6-.1t)^2 + (.9+1.8t)^2)^.5

You can only use the Pythagorean theorem when the terms are sin() and cos() of the same frequency, otherwise they don't correspond properly.

In that expression, you can simplify cos(t)sin(t) using a trig identity so that it works.

Why are you dividing by 2?

 

11 hours ago, djdwosk97 said:

And the frequency of a resonating unforced oscillator would be the same as the frequency of the (non-resonating) unforced oscillator? Because resonance only affects amplitude, right? 

That's correct.

 

11 hours ago, djdwosk97 said:

@Dash Lambda

Sorry to keep asking questions, but I'm taking an online diffeq course, so I don't really have a professor/TA to ask, and I don't seem to have much luck searching for the answers online, and the one textbook I found doesn't really help explain things (at least in a way that I understand them). And your answers do really help me understand what I'm doing.

Happy to help~

Honestly, half the time I check Off-Topic I'm consciously hoping for a math topic XP

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53 minutes ago, Dash Lambda said:

You can only use the Pythagorean theorem when the terms are sin() and cos() of the same frequency, otherwise they don't correspond properly.

In that expression, you can simplify cos(t)sin(t) using a trig identity so that it works.

Why are you dividing by 2?

 

That's correct.

 

Happy to help~

Honestly, half the time I check Off-Topic I'm consciously hoping for a math topic XP

I was dividing by two because I remembered the trig identity incorrectly and dividing by two made it line up with the plot. 

 

haha, well, I'm happy to keep a math topic pinned at the top of off topic :)

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@Dash Lambda

 

How would I find the minimum b such that the amplitude never exceeds 10? I've determined that it's somewhere near .05~ through trial and error, but I assume there is some....more intelligent way of finding the exact number. 

2gCn7KE.png

 

Also a similar problem, is let b=0, what value of A keeps the amplitude <10: f[t]= .5 * Sin[At] where A is as close as possible to Sqrt[2.51]. Again, through trial and error I found it to be around 2.398~, but I assume there is a way that isn't trial and error.

 

One more, for now, would the characteristic equation's solution being +- 2i --> so a period of Pi, be an indication that the plot is close to being in resonance? 

Nq9QfoO.png

 

 

P.s. do you have any textbook recommendations? I looked at Elementary Differential Equations and Boundary Value Problems, and it looks a lot different than what this online course is showing. 

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Sorry for the delayed response, got a little busy with stuff.

 

On 2/25/2017 at 0:57 PM, djdwosk97 said:

How would I find the minimum b such that the amplitude never exceeds 10? I've determined that it's somewhere near .05~ through trial and error, but I assume there is some....more intelligent way of finding the exact number. 

Solving that LDE so the only unknown is b produces the function (The really messy function...):

y = e-bt/2((2 + 1/(2b*sqrt(2.51)))cos(.5sqrt(abs(b2 - 10.04))t) - (2/sqrt(abs(b2 - 10.04)))sin(.5sqrt(abs(b2 - 10.04))t)) - cos(sqrt(2.51)t)/(2bsqrt(2.51))

 

Which has the steady-state (constant amplitude) solution -cos(sqrt(2.51)t)/(2bsqrt(2.51)), which has the amplitude 1/(2bsqrt(2.51)).

The transient (decreasing amplitude) solution influences the oscillation significantly for t<100, but becomes negligible as t increases.

Since the steady-state solution becomes dominant, its amplitude must be less than or equal to 10.

What you need to determine is whether the transient solution constructively interferes (adds to) with the steady-state solution, in which case the amplitude decreases as t increases, or if it destructively interferes (takes away from), in which case the amplitude will increase as t increases (though there may be a 'choke' point), then solve for b accordingly.

 

Funny thing with this one, I solved the LDE at first by representing the circular frequency with a variable to avoid writing it out, but forgot to account for the imaginary number it would produce if abs(b) was too small, so when I substituted the expression back in at the end my solution didn't exist for -sqrt(10.04)<=b<=sqrt(10.04). That confused me for a bit XP

 

On 2/25/2017 at 0:57 PM, djdwosk97 said:

Also a similar problem, is let b=0, what value of A keeps the amplitude <10: f[t]= .5 * Sin[At] where A is as close as possible to Sqrt[2.51]. Again, through trial and error I found it to be around 2.398~, but I assume there is a way that isn't trial and error.

Assuming it's the same DE as the last one, solving: y'' + 2.51y = .5sin(At)

With y(0) = 2, y'(0) = -1

Gives the solution: -sqrt(2.51)(A + 2(2.51 - A2))sin(sqrt(2.51)t)/(2(2.51 - A2)) + 2cos(sqrt(2.51)t) + sin(At)/(2(2.51 - A2))

 

That's two steady-state oscillations with very similar frequencies, which you can simplify algebraically to give you the product of a relatively fast oscillation and a relatively slow oscillation (usually thought of as a fast oscillation with a slowly varying amplitude). From there, you can easily solve for A.

 

On 2/25/2017 at 0:57 PM, djdwosk97 said:

One more, for now, would the characteristic equation's solution being +- 2i --> so a period of Pi, be an indication that the plot is close to being in resonance? 

That question is a little vague, but the answer I'd give is pretty much yours: The roots of the characteristic equation imply that the unforced oscillation has a frequency very close to that of the forcing function.

 

On 2/25/2017 at 0:57 PM, djdwosk97 said:

P.S.: Do you have any textbook recommendations? I looked at Elementary Differential Equations and Boundary Value Problems, and it looks a lot different than what this online course is showing. 

The textbook I have from Diff Eq is this one. Aside from the internet (notable resources being Paul's Math Notes, Khan Academy, Wolfram Mathworld / Alpha, and Wikipedia) and my professor, that's the book I learned from.

 

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Guys I am thinking of going back to university for a computer science degree and I have to learn a lot of new math. 

You guys seem pretty good at it because I understood nothing. My only concern is that i am too dumb to understand the higher math. I bought some books just to understand the basics again and i like it so far but i started with the easy stuff.

 

Any opinion from you guys?

 

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On 2/26/2017 at 7:17 PM, Dash Lambda said:

Sorry for the delayed response, got a little busy with stuff.

 

Solving that LDE so the only unknown is b produces the function (The really messy function...):

y = e-bt/2((2 + 1/(2b*sqrt(2.51)))cos(.5sqrt(abs(b2 - 10.04))t) - (2/sqrt(abs(b2 - 10.04)))sin(.5sqrt(abs(b2 - 10.04))t)) - cos(sqrt(2.51)t)/(2bsqrt(2.51))

 

Which has the steady-state (constant amplitude) solution -cos(sqrt(2.51)t)/(2bsqrt(2.51)), which has the amplitude 1/(2bsqrt(2.51)).

The transient (decreasing amplitude) solution influences the oscillation significantly for t<100, but becomes negligible as t increases.

Since the steady-state solution becomes dominant, its amplitude must be less than or equal to 10.

What you need to determine is whether the transient solution constructively interferes (adds to) with the steady-state solution, in which case the amplitude decreases as t increases, or if it destructively interferes (takes away from), in which case the amplitude will increase as t increases (though there may be a 'choke' point), then solve for b accordingly.

 

Funny thing with this one, I solved the LDE at first by representing the circular frequency with a variable to avoid writing it out, but forgot to account for the imaginary number it would produce if abs(b) was too small, so when I substituted the expression back in at the end my solution didn't exist for -sqrt(10.04)<=b<=sqrt(10.04). That confused me for a bit XP

 

Assuming it's the same DE as the last one, solving: y'' + 2.51y = .5sin(At)

With y(0) = 2, y'(0) = -1

Gives the solution: -sqrt(2.51)(A + 2(2.51 - A2))sin(sqrt(2.51)t)/(2(2.51 - A2)) + 2cos(sqrt(2.51)t) + sin(At)/(2(2.51 - A2))

 

That's two steady-state oscillations with very similar frequencies, which you can simplify algebraically to give you the product of a relatively fast oscillation and a relatively slow oscillation (usually thought of as a fast oscillation with a slowly varying amplitude). From there, you can easily solve for A.

 

That question is a little vague, but the answer I'd give is pretty much yours: The roots of the characteristic equation imply that the unforced oscillation has a frequency very close to that of the forcing function.

 

The textbook I have from Diff Eq is this one. Aside from the internet (notable resources being Paul's Math Notes, Khan Academy, Wolfram Mathworld / Alpha, and Wikipedia) and my professor, that's the book I learned from.

 

I only like half understood that, so I'm still not sure how I'd figure out what b would need to be, but I honestly don't know how well I'd understand that even with more explanation....But you said that the steady-state behavior has an amplitude of: 1/(2*b*sqrt(2.51)), so if b=.0497, then the amplitude would be around 6.35, so why does the plot seem to jump from around 6~ to almost 0 instantly around t=1600~ ? And why does the amplitude it spike up right before dropping to almost 0? 

 

zQvcinL.png

 

And that's the textbook that I've been looking at, but it seems to be worded in a very different way from the way the online course is worded. I'll give it another look. 

 

 

 

Ignore: y=e^(-b*t/2)((2 + 1/(2b*Sqrt[2.51]))Cos[.5Sqrt[Abs[b2 - 10.04]]t] - (2/Sqrt[Abs[b2 - 10.04]])*Sin[.5Sqrt[Abs[b2 - 10.04]]t]) - Cos[Sqrt[2.51]t]/(2b*Sqrt[2.51])

 

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Y U take hard math course, that have no real world use?!

Take a math course that can be use in the real world!

 

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7 hours ago, Kumaresh said:

@NumLock21 WTF are you talking about ?!?!?! D.Es are extremely useful in many fields for research and modelling purposes. I can guarantee you that the practical usefulness of D.Es is far more than something like missile technology......

But missile will let you know in a instant, if you got the equation right or wrong. :P

 

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13 hours ago, Teddy07 said:

Guys I am thinking of going back to university for a computer science degree and I have to learn a lot of new math. 

You guys seem pretty good at it because I understood nothing. My only concern is that i am too dumb to understand the higher math. I bought some books just to understand the basics again and i like it so far but i started with the easy stuff.

 

Any opinion from you guys?

You can understand it, the higher level stuff is essentially just learning how to turn the problems into the basics. It just looks hard because you haven't learned how yet~

 

11 hours ago, djdwosk97 said:

I only like half understood that, so I'm still not sure how I'd figure out what b would need to be, but I honestly don't know how well I'd understand that even with more explanation... But you said that the steady-state behavior has an amplitude of: 1/(2*b*sqrt(2.51)), so if b=.0497, then the amplitude would be around 6.35, so why does the plot seem to jump from around 6~ to almost 0 instantly around t=1600~ ? And why does the amplitude it spike up right before dropping to almost 0? 

That plot is incorrect. It might be an overflow error or something, but that behavior is just too irregular.

Here's a plot of the function and its components with a variable b (the circles next to the functions control visibility): [link]

And for a more in-depth explanation...

 

So this function is the sum of two waves, the transient solution and the steady-state solution, or more specifically the solution with a changing amplitude and the solution with a constant amplitude, respectively.

If b is positive, you can see that the transient solution's amplitude decreases exponentially over time, eventually becoming so small that it can be treated as zero. This means that the way it changes the graph is significant at first, but becomes insignificant as t increases.

At this point, what you need to figure out is how the transient solution changes the graph when it's significant, so you can determine where the function has a maximum amplitude. The easiest way to go about this is graphically:

If you look at the graphs of the component functions (with positive b), you can see that, more or less, one is positive and one is negative. You can think of this as the steady-state solution having a positive amplitude and the transient solution having a negative amplitude, so when you add them together the transient solution decreases the amplitude of the steady-state solution. This means that the amplitude increases over time, approaching the steady-state amplitude as the transient amplitude gets smaller. This means that the maximum is the amplitude of the steady-state solution.

 

One thing to note is that, if you look at the graphs, there is a period where the transient amplitude is larger than the steady-state amplitude. This means that the transient solution is dominant at first, but the steady-state solution becomes dominant once the transient amplitude becomes smaller. This is what causes the 'choke point' at the beginning of the graph.

 

After determining that the maximum amplitude is that of the steady-state solution, you simply need to find for what b that amplitude is 10.

 

11 hours ago, djdwosk97 said:

And that's the textbook that I've been looking at, but it seems to be worded in a very different way from the way the online course is worded. I'll give it another look. 

Oh, I though you were looking at this one...

Well, it's a very good book.

"Do as I say, not as I do."

-Because you actually care if it makes sense.

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13 hours ago, Kumaresh said:

@NumLock21 WTF are you talking about ?!?!?! D.Es are extremely useful in many fields for research and modelling purposes. I can guarantee you that the practical usefulness of D.Es is far more than something like missile technology......

Missle go boom what more do you want or what could be more useful? 

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@Dash Lambda

*riiiiing* *riiiiing*

 

My first question, but I think I figured it out (my understanding is at the top of the spoiler).

 

 


EDIT: From what I've gathered, a plot goes into resonance if its forcing function has some form of cos[z] or sin[z] where z=coefficient of the imaginary of the zsols. So, since z={2i, -2i}, f[t] puts the solution into resonance because the Fourier fit of f[t] has a Cos[2t] term (-2.34735*Cos[2t]). And by making f[t]=f[t]+2.34735*Cos[2t], I can completely get rid of the resonance.

 

How come this goes into resonance? Do you just need to have a term that would put it into resonance (in this case, -2.34375Cos[2t])? And thus if you cancel out that term (by adding 2.34375 to f[t]), then the plot would no longer be in resonance?

WZTiZGf.png
 

 

 

My second question, given a solution, how would I minimize the resonance so that the amplitude doesn't exceed 2 for the first 60 units? And how come simply subtracting the first 1.6t term (1.26596*Cos[1.6t]) doesn't completely eliminate the resonance like it did in the first problem?

TogD3UB.png

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Sorry for the late reply, had some school stuff to sort out.

 

On 3/7/2017 at 1:35 PM, djdwosk97 said:

EDIT: From what I've gathered, a plot goes into resonance if its forcing function has some form of cos[z] or sin[z] where z=coefficient of the imaginary of the zsols. So, since z={2i, -2i}, f[t] puts the solution into resonance because the Fourier fit of f[t] has a Cos[2t] term (-2.34735*Cos[2t]). And by making f[t]=f[t]+2.34735*Cos[2t], I can completely get rid of the resonance.

Correct.

When the solution for the homogeneous equation has the same circular frequency as any of the terms in the forcing function, resonance occurs.

Mathematically, when you solve via undetermined coefficients, your particular solution must be linearly independent of your complementary solution. If terms with the same frequency occur in both, then you multiply that term in the particular solution by t (or whatever other independent variable you're using) to satisfy that independence. That means there will be a term in the solution with an amplitude that varies directly with t, which is responsible for the linear growth in the amplitude of that solution.

 

On 3/7/2017 at 1:35 PM, djdwosk97 said:

My second question, given a solution, how would I minimize the resonance so that the amplitude doesn't exceed 2 for the first 60 units? And how come simply subtracting the first 1.6t term (1.26596*Cos[1.6t]) doesn't completely eliminate the resonance like it did in the first problem?

To completely eliminate resonance, you must remove all terms in the forcing function with the same circular frequency as the solution to the homogeneous equation. This is because any such term must be multiplied by t to make it linearly independent of the complimentary solution.

 

In this situation, if you can only remove one term, you will always have resonance. You can only make it slower.
Any term which is not in resonance only contributes a constant amplitude <=1.5, and therefore cannot mitigate resonance to the degree required. You must find the term that makes the amplitude <=2 at t=60 (because the amplitude is effectively always increasing, nothing before t=60 matters).

I'd need the original equation to see how I'd solve that.

"Do as I say, not as I do."

-Because you actually care if it makes sense.

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12 minutes ago, Dash Lambda said:

To completely eliminate resonance, you must remove all terms in the forcing function with the same circular frequency as the solution to the homogeneous equation. This is because any such term must be multiplied by t to make it linearly independent of the complimentary solution.

So how come removing just 2.343*Cos[2t] eliminated the resonance? Why didn't I have to get rid of the Cos[4t] or Cos[6t] terms? You said that you can't completely eliminate resonance without removing all influencing factors in the forcing function, but for the first part (where I set f[t]=f[t]+2.34375Cos[2t]), the fourier fit of f[t]=1.5624999999999996 - 2.343749999999999*Cos[2*t] + 0.9374999999999996*Cos[4*t] - 0.15625000000000006*Cos[6*t]. 

 

12 minutes ago, Dash Lambda said:

In this situation, if you can only remove one term, you will always have resonance. You can only make it slower.
Any term which is not in resonance only contributes a constant amplitude <=1.5, and therefore cannot mitigate resonance to the degree required. You must find the term that makes the amplitude <=2 at t=60 (because the amplitude is effectively always increasing, nothing before t=60 matters).

I'd need the original equation to see how I'd solve that.

y''[t]+(1.2^2)y[t]=f[t], y[0]=0, y'[0]=0.

f[t]=3*Abs[.2t-Round[.2t]]

fourierfitf[t]=0.7500000000000002 - 0.6114112556352193*Cos[1.2566370614359172*t] - 0.07112944921610617*Cos[3.7699111843077517*t] - 0.028108295546011453*Cos[6.283185307179586*t] - 0.016549903192508993*Cos[8.79645943005142*t] - 0.01220388411722719*Cos[11.309733552923255*t] - 0.010597212292927097*Cos[13.823007675795091*t



o3mtLKt.png

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