Differential Equations Help

[79wjd]

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

[MegaDave91]

[waterjug]

2+2=4

[MegaDave91]

4 minutes ago, wii8cookies said:

2+2=4

Are you positive? Let me count using my fingers.

[DoctorZeus]

at least post a picture of the problem

[79wjd]

3 minutes ago, DoctorZeus said:

at least post a picture of the problem

There's no need for a picture of the problem. It's actually a very simple question for anyone who has taken a Differential Equations class. 

[DoctorZeus]

2 minutes ago, djdwosk97 said:

There's no need for a picture of the problem. It's actually a very simple question for anyone who has taken a Differential Equations class. 

It's so simple? You solved it then! Yay!!

[79wjd]

2 minutes ago, DoctorZeus said:

It's so simple? You solved it then! Yay!!

It's very simple for someone who has taken a diffeq class. I haven't. This is akin to asking how do you find the solutions to something like this: y^2 + 2y + 3 = 0. You just need to know about the quadratic formula and then it's easy. 

[NumLock21]

13 minutes ago, wii8cookies said:

2+2=4

Your calculator is broken!

My is working properly.

1+1=0

[waterjug]

1 minute ago, NumLock21 said:

Your calculator is broken!

1+1=0

oh noes, maybe 1+1=11 because of 42

[DoctorZeus]

2 minutes ago, NumLock21 said:

Your calculator is broken!

My is working properly.

1+1=0

[Codyman125]

As someone who has taken differential equations...

[EunSoo]

i tried, but I didn't understand anything of what you said except for the numbers so I wasn't sure what you are looking for. Im just a high school student with education up to alg2, currently studying trig. This is all I was able to come up with. sorry i couldn't help. good luck tho

[Dash Lambda]

The imaginary part of the root of the auxiliary equation is the circular frequency, and the real part is the coefficient for the exponential.

You basically just have to convert the root from rads/s to revs/s (Hz).

[MrDynamicMan]

7 hours ago, djdwosk97 said:

It's very simple for someone who has taken a diffeq class. I haven't. This is akin to asking how do you find the solutions to something like this: y^2 + 2y + 3 = 0. You just need to know about the quadratic formula and then it's easy. 

Is there really a quadratic formula for *everything? * god damn, and here I am being stumped by adding quad. fractions

[MrDynamicMan]

7 hours ago, SCGazelle said:

i tried, but I didn't understand anything of what you said except for the numbers so I wasn't sure what you are looking for. Im just a high school student with education up to alg2, currently studying trig. This is all I was able to come up with. sorry i couldn't help. good luck tho

My brain bits, they're everywhere. 

[Dash Lambda]

33 minutes ago, MrDynamicMan said:

Is there really a quadratic formula for *everything?* god damn, and here I am being stumped by adding quad. fractions.

In this instance he actually is directly solving a quadratic equation.

[79wjd]

1 hour ago, Dash Lambda said:

 

You basically just have to convert the root from rads/s to revs/s (Hz).

Perfect, thanks.

[79wjd]

@Kumaresh

 

I figured it out thanks to @Dash Lambda 

 

But you can use the characteristic equation in order to solve oscillating diffeq's.

Given: y''[t] + by'[t] + cy[t] = 0, y[0] = x, y'[0] = z

 

Then: 

Characteristic equation  = z^2 + bz + c = 0. (solve for z-> z1, z2) 

GenSol[t] = k1/e^(z1*t) + k2/e^(z2*t) 

GenSol'[t] = z1*t * k1/e^(z1*t) + z2*t * k2/e^(z2*t)

Set gensol[0] = y[0] and gensol'[0] = y'[0] and solve for k1 and k2. 

 

Plug k1/k2 back into gensol[t] and then you can plot that as your oscillating solution to the diffeq. So you can see the frequency in the plot, or you can find the frequency from the characteristic equation. So, according to @Dash Lambda if z = 2Pi i, then the frequency would be 2Pi * 1 Rev/2Pi = 1.

[Blade of Grass]

6 hours ago, MrDynamicMan said:

Is there really a quadratic formula for *everything? * god damn, and here I am being stumped by adding quad. fractions

No! There isn't! For example, there is no known equation to find the roots of a 5^th degree polynomial. 

[79wjd]

13 minutes ago, Kumaresh said:

@djdwosk97 The derivative of gensol[t] looks wrong. Check it again.....

I misstyped typed it, but I wrote it down correctly when I did it. 

 

GenSol'[t] = (-z1 * k1/e^(z1*t)) - (z2 * k2/e^(z2*t))

[Dash Lambda]

9 hours ago, Kumaresh said:

Looks like the DE of a damped oscillation... That is above my area of speciality I don't know all the special forms required to solve these second order DE yet...

It's actually undamped. He gives the general form of a damped oscillation, but the damping coefficient is 0.

 

The characteristic equation of a homogeneous LDE works by assuming a solution can be expressed in the form y = Ce^rt, and since y⁽ⁿ⁾ = rⁿy, you can substitute rⁿ for y⁽ⁿ⁾ to find r. (He used z.)

 

Per this form, if the root is a real number, the solution is an exponential. If it's complex, the solution is y = Ce^{(a + bi)t} = Ce^ate^(bi)t, which by Euler's identity can be expressed as y = e^at(C₁sin(bt) + C₂cos(bt)).

 

Therefore, if the real part is 0, as is the case in x² + c = 0 (where c is a positive constant), e^at = 1 and the motion is undamped.

[Dash Lambda]

53 minutes ago, Kumaresh said:

@Dash Lambda Thanks for the info I didn't pay attention to the coefficients which OP mentioned. I normally use a different method considering it is a homogenous D.E. I thought Euler's identity was e^(ix)=cos(x)+i*sin(x).

That is Euler's identity. Only difference is that the arbitrary coefficients can be complex, so you can make the second coefficient C₂ rather than C₂i.

What method do you normally use?

[79wjd]

On 2/23/2017 at 7:41 AM, Dash Lambda said:

The imaginary part of the root of the auxiliary equation is the circular frequency, and the real part is the coefficient for the exponential.

You basically just have to convert the root from rads/s to revs/s (Hz).

What if z = (-2.85 + 3.46085i) and (-2.85 - 3.46085i)? Would it be: 3.46085/2Pi - 2.85 = 2.3~?

[Dash Lambda]

10 minutes ago, djdwosk97 said:

What if z = (-2.85 + 3.46085i) and (-2.85 - 3.46085i)? Would it be: 3.46085/2Pi - 2.85 = 2.3~?

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 e^ate^(bi)t. When you substitute using Euler's identity, you only replace the exponential containing the imaginary number, so only b influences the frequency.