Differential Equations Help

[79wjd]

@Dash Lambda

Just in case you forgot about me, if you're busy/doing other things, then I can wait

[Dash Lambda]

Sorry, I was helping some friends move and didn't really have time. Done now though.

 

On 3/9/2017 at 7:32 PM, djdwosk97 said:

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]. 

You have to remove all components with the same circular frequency as the complementary solution, so in the first one you only needed to remove all '2t' terms. In the second one, there is both a sin(1.6t) and a cos(1.6t).

 

On 3/9/2017 at 7:32 PM, djdwosk97 said:

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

Hmm... Where'd the 1.6 from before come from?

In that problem, the complementary solution has a circular frequency of 1.2, so you need to remove from the forcing function every term with a circular frequency that is approximately 1.2 (it can cause resonance by being close even if not equal), which would be the single ~cos(1.2566t) term.

[79wjd]

24 minutes ago, Dash Lambda said:

Sorry, I was helping some friends move and didn't really have time. Done now though.

 

You have to remove all components with the same circular frequency as the complementary solution, so in the first one you only needed to remove all '2t' terms. In the second one, there is both a sin(1.6t) and a cos(1.6t).

 

Hmm... Where'd the 1.6 from before come from?

In that problem, the complementary solution has a circular frequency of 1.2, so you need to remove from the forcing function every term with a circular frequency that is approximately 1.2 (it can cause resonance by being close even if not equal), which would be the single ~cos(1.2566t) term.

Ah okay that explains it, I must've done the math wrong somehow and got 1.6, no wonder I wasnt able to cancel out the resonance even when I also removed the Sin[1.6] term.

[79wjd]

@Dash Lambda

Edit: My bad, there were two problems and I got confused with my "correction posts". 

 

Problem #1: Add an extra term of the form A Cos[B t] or A Sin[B t] with -0.5 <= A <= 0.5 to f[t]?" Obviously, if I just do f[t]=f[t]+.611411*Cos[1.2566t] it will basically be taken care of, but I need to A to be <.5, I thought of just using a trig identity to convert it, but I couldn't think of one that works (so I don't know if I'm just overlooking something or if there is another way to do it -- I used Sin^2(t)+Cos^2(t)=1 and Sin^2(t)=Cos^2(t)+Cos(2t)).

 

y''[t]+1.2^2y[t]=f[t].

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

y[0]=0, y'[0]=0.

fastfourierfitf[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
 

 

 

 

Problem #2: Add an extra term of the form A*Cos[bt] or A*Sin[bt] with -1.5<=A<=1.5 to f[t] in order to keep the amplitude <=2 for 0<t<60.

 

 

y''[t]+1.6^2y[t]=f[t].

 

f[t]=Sign[Cos[0.5 Pi t]] + 1

fastfourierf[t]=(1.0000000000000002 + 0.*I) + 1.2659590187875254*Cos[1.6000003328706176*t] - 0.4023689270621826*Cos[4.800000998611853*t] + 0.21720422880686774*Cos[8.00000166435309*t] - 0.12788783132982676*Cos[11.200002330094323*t] + 0.0690355937288492*Cos[14.400002995835559*t] - 0.02194208293123261*Cos[17.600003661576796*t] + 0.1666666666666667*Sin[1.6000003328706176*t] - 0.1666666666666667*Sin[4.800000998611853*t] + 0.1666666666666667*Sin[8.00000166435309*t] - 0.16666666666666669*Sin[11.200002330094323*t] + 0.1666666666666667*Sin[14.400002995835559*t] - 0.16666666666666666*Sin[17.600003661576796*t
 

 

[Dash Lambda]

On 3/12/2017 at 10:45 PM, djdwosk97 said:

Problem #1: Add an extra term of the form A Cos[B t] or A Sin[B t] with -0.5 <= A <= 0.5 to f[t]?" Obviously, if I just do f[t]=f[t]+.611411*Cos[1.2566t] it will basically be taken care of, but I need to A to be <.5, I thought of just using a trig identity to convert it, but I couldn't think of one that works (so I don't know if I'm just overlooking something or if there is another way to do it -- I used Sin^2(t)+Cos^2(t)=1 and Sin^2(t)=Cos^2(t)+Cos(2t)).

  Reveal hidden contents

y''[t]+1.2^2y[t]=f[t].

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

y[0]=0, y'[0]=0.

fastfourierfitf[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
 

 

 

If the goal is to keep the amplitude within a given maximum for a given range, I'd how much you need to reduce that term's coefficient to achieve that.

 

On 3/12/2017 at 10:45 PM, djdwosk97 said:

Problem #2: Add an extra term of the form A*Cos[bt] or A*Sin[bt] with -1.5<=A<=1.5 to f[t] in order to keep the amplitude <=2 for 0<t<60.

 

  Hide contents

y''[t]+1.6^2y[t]=f[t].

 

f[t]=Sign[Cos[0.5 Pi t]] + 1

fastfourierf[t]=(1.0000000000000002 + 0.*I) + 1.2659590187875254*Cos[1.6000003328706176*t] - 0.4023689270621826*Cos[4.800000998611853*t] + 0.21720422880686774*Cos[8.00000166435309*t] - 0.12788783132982676*Cos[11.200002330094323*t] + 0.0690355937288492*Cos[14.400002995835559*t] - 0.02194208293123261*Cos[17.600003661576796*t] + 0.1666666666666667*Sin[1.6000003328706176*t] - 0.1666666666666667*Sin[4.800000998611853*t] + 0.1666666666666667*Sin[8.00000166435309*t] - 0.16666666666666669*Sin[11.200002330094323*t] + 0.1666666666666667*Sin[14.400002995835559*t] - 0.16666666666666666*Sin[17.600003661576796*t
 

 

Instead of looking at the Fourier fit, you should look at the original forcing function and see if you can mitigate the resonance enough by reducing the amplitude of the overall wave (doesn't have to stay a square wave) to fit within the constraint.

Are there initial conditions?

[79wjd]

25 minutes ago, Dash Lambda said:

If the goal is to keep the amplitude within a given maximum for a given range, I'd how much you need to reduce that term's coefficient to achieve that.

 

Instead of looking at the Fourier fit, you should look at the original forcing function and see if you can mitigate the resonance enough by reducing the amplitude of the overall wave (doesn't have to stay a square wave) to fit within the constraint.

Are there initial conditions?

For the first problem, the amplitude needs to remain <=4 (it doesn't give a time window).

 

For the second problem, y[0]=0, y'[0]=0. Also, I think the point of this exercise is to use the fourier fit in order to reduce the resonance (also, I have no idea how to do it by just looking at the forcing function).

[Dash Lambda]

On 3/15/2017 at 10:07 PM, djdwosk97 said:

For the first problem, the amplitude needs to remain <=4 (it doesn't give a time window).

Then the question is incomplete. Even if it's supposed to always stay within that amplitude, it should say something like "at all times."

You can keep the amplitude within 4 for an initial range if you remove as much of the relevant term as possible (without removing all of it), but you can't completely eliminate the resonance with the requirements they give you.

 

On 3/15/2017 at 10:07 PM, djdwosk97 said:

For the second problem, y[0]=0, y'[0]=0. Also, I think the point of this exercise is to use the fourier fit in order to reduce the resonance (also, I have no idea how to do it by just looking at the forcing function).

The wave of the forcing function has a period of 4, so you could reduce the influence of the overall wave by removing a corresponding simple wave (rather than components).

I checked, though, and that can't solve the problem, so never mind on that...

 

In order to solve that, I'd try to subtract a term of the form Asin(Bt - a), but I don't know how I'd solve it with the given options.

[79wjd]

@Dash Lambda I'm supposed to use separation of variables to determine the diffeq and then determine if it "escapes to infinity in an infinite time or a finite time", but I'm not sure how to do that last bit. 

 

Here's an example of the problem:

[Dash Lambda]

On 4/3/2017 at 7:46 PM, djdwosk97 said:

@Dash Lambda I must be overthinking this....but, I have to use "separation of variables" in order to solve: y'[t]=1+y[t]/2. Separation of variables being that I'm supposed to get y'[t] and y[t] on the same side and the t terms (which aren't in this particular problem?) to the other side.

 

Here's the example problem:

[image]

But if I follow the example, then I end up trying to take the integral of y[x], which doesn't work: 

[image]

And solving it the normal way is trivially easy, but that's not the point of this question. 

I think part of the problem is that they're trying to teach it without altering the notation. For separation of variables, it's a lot easier if you write the derivative as dy/dt, as opposed to y'(t). The goal isn't necessarily to isolate all the y terms on one side and all the t terms on the other, it's to isolate them into separate integrable expressions, which when you use dy/dt means putting it into the form ([expressionY])dy = ([expressionT])dt.

 

So, essentially, rather than adding and subtracting terms of y, multiply and divide expressions containing terms of y.

 

EDIT: Noticed right after posting that you'd figured it out, give me a minute. Sorry, started this response right after you posted but I've been quite busy the last few days >.<

[79wjd]

Just now, Dash Lambda said:

I think part of the problem is that they're trying to teach it without altering the notation. For separation of variables, it's a lot easier if you write the derivative as dy/dt, as opposed to y'(t). The goal isn't necessarily to isolate all the y terms on one side and all the t terms on the other, it's to isolate them into separate integrable expressions, which when you use dy/dt means putting it into the form ([expressionY])dy = ([expressionT])dt.

 

So, essentially, rather than adding and subtracting terms of y, multiply and divide expressions containing terms of y.

Yeah, I figured that out and edited my post a couple days ago

 

I'm still not sure how to determine if it escapes to infinity in infinite time or finite time. 

[Dash Lambda]

10 minutes ago, djdwosk97 said:

Yeah, I figured that out and edited my post a couple days ago

 

I'm still not sure how to determine if it escapes to infinity in infinite time or finite time. 

Yeah, I edited my response with an explanation, sorry about that XP

 

I think what they're asking is if the solution increases unbounded at a point, like 1/t is at t=0, or if it increases unbounded along t. The language is a little odd, but that seems like the most reasonable interpretation.

EDIT: Most of the questions like this I encountered were asked (roughly) in the form: "Is the function continuous along ARN? If not, explain the nature of the discontinuity."

[bcguru9384]

On ‎2‎/‎22‎/‎2017 at 10:48 PM, djdwosk97 said:

How do you find the frequency of an oscillation from the characteristic equation without solving the whole diffeq? 

b = 0

c = 4 pi²

z² + bz + c = 0, z = +- 2i Pi

((±2iπ)(±2iπ))+((0)(±2iπ))+(4π²)=0

either your 0 result is wrong or your z² not z² but is z*z

unless

(((2(±i)π))((2(±i)π)))

now i gets to be variably different and will allow a -4π² cancel +4π²

i=1 the only changeable part is ± you must have both 

-1 && +1

i think 

Edited April 7, 2017 by bcguru9384
added reason

[NumLock21]

My god, this problem still isn't solve yet?!

 

 

[bcguru9384]

now that we know (i) must equal 1

we can build i equations however we want to as long as we have send and receive equation

i positive 1 will be send

i negative 1 is receive 

equation then may confirm a 0(zero) loss

or request the missing parts

no matter how much data to be sent

[AniJan]

I only know the ALGEBRA 1...

[79wjd]

11 hours ago, NumLock21 said:

My god, this problem still isn't solve yet?!

 

 

It is, I've just been using this thread to ask @Dash Lambda new questions.

11 hours ago, bcguru9384 said:

now that we know (i) must equal 1

we can build i equations however we want to as long as we have send and receive equation

i positive 1 will be send

i negative 1 is receive 

equation then may confirm a 0(zero) loss

or request the missing parts

no matter how much data to be sent

What are you even saying...

10 hours ago, AniJan said:

I only know the ALGEBRA 1...

I miss algebra 1.

12 hours ago, Dash Lambda said:

Yeah, I edited my response with an explanation, sorry about that XP

 

I think what they're asking is if the solution increases unbounded at a point, like 1/t is at t=0, or if it increases unbounded along t. The language is a little odd, but that seems like the most reasonable interpretation.

EDIT: Most of the questions like this I encountered were asked (roughly) in the form: "Is the function continuous along ARN? If not, explain the nature of the discontinuity."

I found this when I try googling the problem.... But it just confuses me more: http://math.stackexchange.com/questions/1163591/solutions-escape-to-infinity-in-finite-time

How does 1/X[sub0] escape to infinity in finite time, but e^t takes infinite time?

[Dash Lambda]

1 hour ago, djdwosk97 said:

I found this when I try googling the problem.... But it just confuses me more: http://math.stackexchange.com/questions/1163591/solutions-escape-to-infinity-in-finite-time

How does 1/X[sub0] escape to infinity in finite time, but e^t takes infinite time?

1/(1/x₀-t) evaluates to 1/0 when t=1/x₀, meaning the limit at the finite number 1/x₀ grows without bound.
Exponentials, however, are continuous and differentiable across the entire real number line (with a positive base), meaning that they don't grow unbounded at any finite point, but they are always increasing with time, meaning that as t approaches infinity the function does as well.

[bcguru9384]

as always with us humans

we steer away from simple

and 

we look for reasons to argue

it is an alegbra problem

π who used the entire number in their solution??? no one. 

[79wjd]

5 minutes ago, bcguru9384 said:

as always with us humans

we steer away from simple

and 

we look for reasons to argue

it is an alegbra problem

π who used the entire number in their solution??? no one. 

Yeah.....I don't think you have a clue what you're talking about.

[bcguru9384]

yes i guess your right 

goodbye

[79wjd]

@Dash Lambda

 

I've ran into a couple questions that are stumping me.

 

Question 1: 

 

Question #2:

[Dash Lambda]

1: For the line to go with the flow, it must be a solution to the DE. The given (incorrect) answer is a superposition of only the vectors, not of solutions.

 

2: A linear system is linear with respect to each function and its derivatives.
Taking the first one as example:
sin(x)+4.9y is nonlinear with respect to x as sin is a nonlinear function.
-2.3x+.4y is linear.
Because part of the system is nonlinear, the system is nonlinear.

 

EDIT: Sorry again for the wait, I hate taking so long.

[bcguru9384]

On ‎5‎/‎5‎/‎2017 at 2:56 PM, djdwosk97 said:

Question #2:

(your entire equation)=1(second)÷#ofsteps pretend that (1/(#)=(y=mx+b))=times involvement 

(()=(1/60))

[Dash Lambda]

Just now, bcguru9384 said:

(your entire equation)=1(second)÷#ofsteps pretend that (1/(#)=(y=mx+b))=times involvement 

(()=(1/60))

... This confuses me.

[79wjd]

16 minutes ago, Dash Lambda said:

1: For the line to go with the flow, it must be a solution to the DE. The given (incorrect) answer is a superposition of only the vectors, not of solutions.

 

2: A linear system is linear with respect to each function and its derivatives.
Taking the first one as example:
sin(x)+4.9y is nonlinear with respect to x as sin is a nonlinear function.
-2.3x+.4y is linear.
Because part of the system is nonlinear, the system is nonlinear.

I get #2, but I'm a bit confused on #1. How would I correct it then, how would I solve the diffeq without starter data?