2nd order differential equation

[kristiakemi]

I am currently solving a physics problem that requires me to solve the following equation

m(d^2x/dt)=-gamma-c(dx/dt) but I can't seem to come up with a method that makes sense. Note: Gamma is just some constant. I tried to integrate both sides wrt t but then I end up with both a velocity function and x(t) in the same equation and that physically doesn't make sense.

I've also tried to turn this into a 2nd order differential equation by saying:

m(d^2x/dt)+c(dx/dt)+gamma=0 . However, now, the issue is that I can't turn this into the characteristic polynomial because gamma does not have an x.

The gist of the problem is that For what Vo must a cup experience in order for it to fall off of a table of length H if the only force acting on the cup is coulumb friction, -gamma. I found that Vo>sqrt(2Hgamma/M). Now I need to solve this same problem when coulomb friction is now -gamma-c(dx/dt).

Any help solving m(d^2x/dt)+c(dx/dt)+gamma=0 or an idea of a different method would be great. thanks!

 

[arildno]

Well, just invent the function u(t)=dx/dt to convert your equation into a first order linear for u, and use, for example, an integrating factor to solve for u.
Then solve for x(t).

 

[HallsofIvy]

Or, conversely, write it as the non-homogeneous linear differential equation with constant coefficients, md^2x/dt^2+ c dx/dt= \gamma. Solve the related homogenous equation and use "undetermined coefficients" to find a specific solution to the entire equation.

 

[Matthew Rodman]

I'd use arildno's method if I were you. Set u= dx/dt, and your equation becomes

u^{\prime} + \frac{c}{m} u + \frac{\gamma}{m} = 0

Now, the solution of this is reasonably straightforward. If m,c and gamma are all constants, you can see that the solution for u(t) is of the form

u(t) = Ae^{-\frac{c}{m} t} + B

where A,B are constants. To determine B, you can differentiate this and plug it into the u equation, and you'll see that

B = -\frac{\gamma}{c}

Therefore,

u(t) = Ae^{-\frac{c}{m} t} - \frac{\gamma}{c}

and you only need to integrate to find x(t). Also - if you have boundary conditions, you'll have to use those to find the value of A ( and the constant of integration that will emerge when you integrate u(t) ).

Are m and c constants? If not, you'll need a more general way of solving the u equation.

 

[HallsofIvy]

I'd use arildno's method if I were you.

I wouldn't! It's a matter of personal taste.

The characteristic equation of the homogeneous, linear equation with constant coefficients,
m\frac{d^2x}{dt}+ c\frac{dx}{dt}= 0
is r^2+ cr= r(mr+ c)= 0 which has roots 0 and -c/m. That tells us that the general solution to the homogeneous equation is C+ De^{-ct/m} for constants C and D.

Because the right hand side of the entire equation is a constant, we would normally try a 'specific solution' also a constant- however, since constant C is a solution to the homogeneous equaton, we try instead x= At. Then x'= A and x"= 0. The equation become cA= -\gamma so A= -\gamma/c.

The general solution to the differential equation is
x(t)= C+ De^{-\frac{ct}{m}}- \frac{\gamma}{c}t

 

[Plastic Photon]

divide by m to get a constant coeffecient equation, then solve for the reduced equation d^2x/dt^2 + (c/m)dx/dt = 0 and then solve for gamma/m using D as a constant

 

[HallsofIvy]

?? It is a "constant coefficient equation" to begin with. Dividing by m doesn't change that!

 

[D H]

You can also solve this as a vector equation. In this case, the vector form is rather tedious. It can be quite useful (particularly if one has to resort to numerical methods to solve the differential equation).

Invent a vector variable u such that
\vec u=\bmatrix x\\ \dot x\endbmatrix

Then

\frac {d \vec u}{dt} =<br /> \bmatrix 0&amp;1\\0&amp;-\frac c m\endbmatrix\; \vec u +<br /> \bmatrix 0\\ -\frac{\gamma}m\endbmatrix

Let
\mathbf A = \bmatrix 0&amp;1\\0&amp;-\frac c m\endbmatrix

\vec b = \bmatrix 0\\ -\frac{\gamma}m\endbmatrix

making

\frac {d \vec u}{dt} = \mathbf A\vec u + \vec b

The homogeneous solution

\frac {d \vec u}{dt} = \mathbf A\vec u

has solution

\vec u = \exp(\mathbf A t) \bmatrix x_h \\ v_h \endbmatrix

where x_h and v_h are arbitrary constants and

\exp(\mathbf A t)<br /> \equiv \mathbf I + \sum_{n=1}^{\infty}\frac{t^n}{n!}\mathbf A^n

In this case,

\mathbf A^n = \left(-\frac c m\right)^{n-1} \mathbf A, n \ge 1

and thus

\exp(\mathbf A t) = \mathbf I + \frac m c(1-e^{-\frac c m t})\mathbf A

making the homogeneous solution

\vec u_h =<br /> \bmatrix<br /> x_h + \frac m c(1-e^{-\frac c m t}) v_h \\<br /> e^{-\frac c m t} v_h<br /> \endbmatrix

A specific solution to the imhomogeneous equation is

\vec u_s = -\frac{\gamma}c\bmatrix t\\ 1\endbmatrix

Combining the homogeneous and imhomogeneous solutions,

\vec u = <br /> \bmatrix<br /> x_h + \frac m c(1-e^{-\frac c m t}) v_h - \frac{\gamma}c t\\<br /> e^{-\frac c m t} v_h -\frac{\gamma}c<br /> \endbmatrix

The solution is thus

x(t) = x_h + \frac m c(1-e^{-\frac c m t}) v_h - \frac{\gamma}c t