Question

mx''+kx=Acos(pt)

Is there a general solution to this, not just where there is resonance or similar?

Answer 1

I don't think I get your terminology, but the form of the answer will depend on the sign of k, which is I assume your damper/amplifier

Answer 2

Forced oscillator.

Answer 3

like all second-order non-homogeneous DE's it will have the form\[y=y_c+y_p=c_1y_1+c_2y_2+y_P\]and \(y_c\) depends on the value of k being positive, negative, or zero

Answer 4

Or, to go for the whole hog, is there an exact solution for\[m\frac{d^2x}{dt^2}=-kx-c\frac{dx}{dt}+acos(\theta t)\]

Answer 5

@experimentX , do you have any ideas how to tackle the second DE? Despite both of your help, I'm still very much guessing solutions to DE based on aforeknown schemes.

Answer 6

this is your homogeneous DE.
\[ m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = 0\]
do you know how to find it's solution?

Answer 7

Assume \[x=Ae^{pt}\]Divide by\[Ae^{pt}\]Solve the quadratic for p and find A by initial conditions.

Answer 8

yep ... it can be modelled as harmonic oscillator
c/m as your damping constant and sqrt(k/m) as your oscillating frequency

Answer 9

\[ m\frac{d^2x}{dt^2} + c\frac{dx}{dt} + kx = f(t) \]
and f(t) is your input to the system. the solution would be sum of complimentary function (which is the solution of homogeneous DE) and Particular Integral.

we add complimentary function because you can test it in DE. i doesn't affect the result.

Answer 10

And also I'll give you one basic intuition behind \( A e^{pt } \).
you can consider is as linear operator on x
\[ L(D) x = (D-a)(D-b)y = 0 --(1) \\
\]
Let
\[ (D- b)y = u ---(2)\]
then we have
\[ (D-a)u = 0 --- (3)\]
this is linear in u ... solve for u, then put it in (2) and solve whole DE. probably you willget in terms of e^(px) is is characteristic of first order DE.

Answer 11

solving for particular integral is not quite easy. there are bunch of techniques. most used are
1. undetermined coefficients (this is purely guessing)
2. reduction of orders (i like this one)
3. variation of parameters
4. and there is one commonly used ones here (in my place) ...perhaps you guys are not familiar with it.

Answer 12

Before I read your last & penultimate posts: how do we know that adding particular and complementary is enough? Couldn't there be a huge number of complimentary solutions 'hiding'?

Answer 13

no there cannot be more that two integral constants. that is from the definition of DE.
DE are made by eliminating these constants.

Answer 14

if you have simple input type of polynomials or exponential or trigonometric ... consider yourself lucky.
polynomials >> you can easily guess solution
exponential input >> http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/exponential-response/
this is easy to remember
trigonometric >> can be changed into exponential using Euler's formula. just look for imaginary part (if you have sine) or real part ... if you have cosine.

Answer 15

What are the commonly used ones for you?

Answer 16

And thanks for the explanation and link, twas verily useful.

Answer 17

i don't understand how they do it ... but's it's like using
\[ {1 \over D} x = \int x dx\]
I'll show you for particular problem
\[ (D - 1)(D - 2)x = t \\
P.I. = {1 \over (D - 1)(D - 2)} \;t = {1 \over -2( D - 1)} (1 + {D \over 2} + {D^4 \over 4} + ..)t =\\ -{1 \over 2 (D -1)}(t + 1/2) = {1 \over 2} ( 1 +D + D^2 + ..)(t+ 1/2) = 1/2(t + 1/2) + 1/2\]

something like this ... if you encounter 1/D ... you integrate.

Answer 18

in English this is the annihilator method

Answer 19

hmm ... really? i didn't know that. Man I'm really bad at English.

Answer 20

Thanks. And sorry- I'm not familiar with the D notation. D is d/dx?

Answer 21

yep ... this acts as linear operator.

Answer 22

no you're not, it's just a piece of terminology
If we are in fact talking about the same thing D is the "differential operator" and acts like taking a derivative.

Answer 23

I'm not so savvy on the technique myself

Answer 24

Dy =d(y)/dx
L(D)y= L(d(y)/dx)?

Answer 25

I only learnt about operators the other day, so pardon my ineptitude.

Answer 26

this method gives quick and dirty solution.

Answer 27

L(D) = L(d/dx) ... not dy/dx ... just d/dx .... this will operate on 'y'

Answer 28

L(d/dx) is just some function of d/dx

Answer 29

So functions are written as F(y) , and operators as Fy?

Answer 30

No ... L(d/dx) whole is operator
\( {d^2y \over dx^2} + 2{dy \over dx} + y = ({d^2 \over dx^2} +2 {d \over dx} + 1) y = (D^2 + 2D + 1)y = F(D) y\)

Answer 31

you choose for instance the operator\[(D^2+4)\]to annihilate\[\sin(2x)\] because\[(D^2+4)\sin(2x)=D^2(\sin(2x))+4\sin(2x)\]\[=D(2\cos(2x))+4\sin(2x)=-4\sin(2x)+4\sin(2x)=0\]hence the name

Answer 32

I wrote L in-place of F ... because it is linear.

Answer 33

\[\frac{1}{D}x=\int\limits x dx\] So\[\frac{1}{F}x=y \rightarrow Fy=x\] So is an inverse operator exactly the same as an inverse function? Does the 1/D notation imply that you can algebraically manipulate an equation with D as if it were a constant?

Answer 34

well ... sure you can manipulate D's to your ease. but careful when dealing with variables.
For example in above ... you see i manipulated it as geometric series and got the particular integral. http://www.wolframalpha.com/input/?i=y%27%27+-+3+y%27+%2B+2y+%3D+x
similarly you can try this
\[ D(D-1)y = x\]

Answer 35

\[D(D-1)y=\frac{dy}{dx}(\frac{dy}{dx}-1)?\]

Answer 36

Is this a correct reading of the notation?

Answer 37

Oops sorry no ...
\[ D(D-1)y=(\frac{d}{dx}(\frac{d}{dx}-1) )y \]

Answer 38

What does that MEAN, though? What do you do to y, and in what order?

Answer 39

Damn ... chrome just crashed ... therefor no latex.