# 2nd order differential equation

• kristiakemi
In summary, the conversation discusses a physics problem involving a second order differential equation m(d^2x/dt)+c(dx/dt)+gamma=0 and the methods used to solve it. The conversation suggests using either a first order linear differential equation or a non-homogeneous linear differential equation with constant coefficients to solve for the function u(t)=dx/dt. The solution for x(t) can then be found by integrating. Another method mentioned is to solve the equation as a vector equation using

#### 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!

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).

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.

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.

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

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

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

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

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

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## 1. What is a 2nd order differential equation?

A 2nd order differential equation is a mathematical equation that involves a function and its first and second derivatives. It represents the relationship between the function and its rate of change.

## 2. What is the general form of a 2nd order differential equation?

The general form of a 2nd order differential equation is y'' + p(x)y' + q(x)y = g(x), where y is the function, p(x) and q(x) are coefficients, and g(x) is the source or forcing function.

## 3. What are the types of solutions for a 2nd order differential equation?

The types of solutions for a 2nd order differential equation depend on the values of p(x) and q(x). If p(x) = 0 and q(x) = 0, the equation has a general solution of y = C1x + C2, where C1 and C2 are constants. If p(x) and q(x) are both non-zero, the equation has a unique solution. If p(x) and q(x) are both zero, the equation has an infinite number of solutions.

## 4. How are 2nd order differential equations solved?

2nd order differential equations can be solved using various methods such as separation of variables, variation of parameters, and Laplace transforms. The specific method used depends on the type of equation and its initial or boundary conditions.

## 5. What are the applications of 2nd order differential equations?

2nd order differential equations have many applications in physics, engineering, and other sciences. They are used to model and analyze systems that involve acceleration, oscillations, and other types of motion. Some examples include the motion of a pendulum, the growth of a population, and the behavior of electrical circuits.