# How to integrate second order derivative

1. May 14, 2014

### Vibhor

I would like to know how do we solve d2x/dt2 = k' where k' is a constant i.e the task is to find x as a function of time ?

One way to approach this is to rewrite it as vdv/dx = k' where v=dx/dt and first find find v as a function of x and then rewrite v as dx/dt and then find x as a function of time .

I will present my attempt .

vdv/dx = k'
vdv= k'dx
∫vdv= ∫k'dx
v2 = 2k'x + 2C' where C' is a constant.

v=√(kx+C)
Now,v=dx/dt

dx/√(kx+C) =dt
∫dx/√(kx+C) =∫dt

x = (αt+β)2 ,where α and β are some constants.

Now ,I would like to know how do we solve the equation d2x/dt2 = k' by direct integration.

∫(d2x/dt2)dt = ∫k'dt

How to integrate the left hand side as it is a second order derivative ?

Last edited: May 14, 2014
2. May 14, 2014

### D H

Staff Emeritus
Rhetorical question: Is this result correct?

When you differentiate the above twice with respect to time you should get $\frac{d^2x(t)}{dt^2} = ke^{kt+\lambda}$. That obviously is not correct. Moral of the story: Always double check your work.

You can arrive at the correct expression using the fact that $\frac{d^2x(t)}{dt^2} = v\frac{dv}{dx}$. However, this is a much more difficult approach than it is to first integrate $\frac{dv}{dt} = k$ directly and then integrating that result with respect to time to arrive at an expression for x(t). Both of these integrals are very easy.

3. May 14, 2014

### Vibhor

No .

I apologize for the silly mistake .I have edited my post.

OK .Thanks !

You are right .But I would like to know do we always have to go by this two step process or is it possible to deal with $\int \frac{d^2x(t)}{dt^2}dt$ directly .

4. May 15, 2014

### csleong

I am very confused with the question, for me isn't it a simple answer as x=0.5k't^2 + at + b?

5. May 20, 2014

### sudhirking

I would like to answer a slightly different question. Suppose you had to integrate the first derivative of a function. How then would you proceed?

Your response would be something along the lines of, "Well of course FTC! The integral of a derivative would be up to some constant, the function itself".

To call something a second derivative is, in this circumstance, superficial since that second derivative can be written as the first derivative of some other function (namely, the first derivative of the original).

With this substitution, carry out your integration using FTC. Now you will have solved for not the original function but for its derivative. Integrate yet again. Remember to keep track of your constants.