# 2nd order differential equation

1. Feb 17, 2007

### 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 doesnt 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!!

2. Feb 18, 2007

### 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).

3. Feb 18, 2007

### 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.

4. Feb 19, 2007

### 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.

Last edited: Feb 19, 2007
5. Feb 20, 2007

### HallsofIvy

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
$$x(t)= C+ De^{-\frac{ct}{m}}- \frac{\gamma}{c}t$$

6. Feb 20, 2007

### 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

7. Feb 23, 2007

### HallsofIvy

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

8. Feb 23, 2007

### D H

Staff Emeritus
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

$$\frac {d \vec u}{dt} = \bmatrix 0&1\\0&-\frac c m\endbmatrix\; \vec u + \bmatrix 0\\ -\frac{\gamma}m\endbmatrix$$

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

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

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

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

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

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

A specific solution to the imhomogeneous equation is

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

Combining the homogeneous and imhomogeneous solutions,

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

The solution is thus

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

Last edited: Feb 23, 2007