# How to find the constant in this indefinite integration?

Istiak
Homework Statement:
Find constant of a indefinite integration for classical mechanics
Relevant Equations:
$\int$
$$x(t)=\int \dot{x}(t)\mathrm dt=vt+c$$

That's what I did. But, book says

$$x(t)=\int \dot{x}(t)\mathrm dt=x_0+v_0 t+ \frac{F_0}{2m}t^2$$

Seems like, $$x_0 + \dfrac{a_0}{2}t^2$$ is constant. How to find constant is equal to what?

Mentor
2021 Award
To find the value of the constant, you will need a initial condition, often ##x(0)=0## or in general ##x(t_0)=x_0##. Then insert it in your equation and solve for ##c##.

The reason is the following:
A differential equation has - depending on c - many solutions. Each solution corresponds to a flow through a vector field which is defined by the differential equation. The initial condition tells us where to start the flow, such that it will be the unique one that solves the equation plus its initial condition.

ref.: https://mathematica.stackexchange.com/questions/22190/why-are-these-flow-lines-cut-short

Mentor
What is ##\dot{x}(t)## as a function of t? Don't assume it's a constant.

Istiak
To find the value of the constant, you will need a initial condition, often x(0)=0 or in general x(t0)=c0. Then insert it in your equation and solve for c.
If I try to do this way then, I get c=-vt

0=vt+c
c=-vt

Mentor
is velocity... It's obviously constant [as far as I know]
Why do you assume that? Your book didn't!

Istiak
Why do you assume that? Your book didn't!
What my book didn't do?

Mentor
What my book didn't do?
It looks to me as if the problem assumes a constant acceleration, not a constant velocity.

Mentor
2021 Award
If I try to do this way then, I get c=-vt

0=vt+c
c=-vt
If you set ##t=0## then you get ##c=x(0)## in the first equation, and ##x(0)=x_0## in the second.

Initial conditions could as well be at any time and of any value, e.g. ##x(5)=7## but it is usually ##x(0)## we are interested in because it makes the equations easier to solve for ##c##. It is also where the name comes from: initial as in starting point which is usually ##t=0.##

Istiak
Istiak
If you set ##t=0## then you get ##c=x(0)## in the first equation, and ##x(0)=x_0## in the second.

Initial conditions could as well be at any time and of any value, e.g. ##x(5)=7## but it is usually ##x(0)## we are interested in because it makes the equations easier to solve for ##c##. It is also where the name comes from: initial as in starting point which is usually ##t=0.##
So, I can find $$x(t)=vt+x(0)$$ but, my book assumes that $$x(t)=x(0)+vt+\frac{at^2}{2}$$

Mentor
So, I can find $$x(t)=vt+x(0)$$ but, my book assumes that $$x(t)=x(0)+vt+\frac{at^2}{2}$$
You assume that the velocity is constant, but your book does not.

Delta2
Istiak
You assume that the velocity is constant, but your book does not.
How? If I integrate velocity than, I get position not acceleration...

Mentor
How? If I integrate velocity than, I get position not acceleration...
You're trying to find the position.

Do this: Start with a constant acceleration. Then integrate to find the velocity, and then integrate again to find the position.

Istiak
Mentor
2021 Award
So, I can find $$x(t)=vt+x(0)$$ but, my book assumes that $$x(t)=x(0)+vt+\frac{at^2}{2}$$
So? Your book says ##x(0)=v\cdot 0 +x(0)## in the first and ##x(0)=x(0) + v\cdot 0 +\frac{a\cdot 0^2}{2}=x(0)## in the second equation. But what does it say in the problem statement before ##c## occurs?

Without initial condition ##x(t_0)=vt_0+x_0## you get infinitely many solution from ##\dot x =v## and with initial condition only one solution: ##x(t)=v\cdot t +x_0##.

And in order to solve ##\ddot x = a## you need even two initial conditions because you integrate twice and each integral has a ##c##.

Istiak
Mentor
I got confused... $$\dot{x}=\int \ddot{x} \mathrm dt=\dot{x} t +c$$
That's a good start. Now solve for that constant. Looks like they assume the velocity is v0 when t = 0.

Istiak
You're trying to find the position.

Do this: Start with a constant acceleration. Then integrate to find the velocity, and then integrate again to find the position.
Ohh! Thanks... got it...

$$\dot{x} (t)=\int \ddot{x} t dt=\ddot{x}t+c$$
$$\dot{x}(t)=\ddot{x}t+\dot{x}(0)$$
$$x(t)=\int \dot{x} (t) dt$$
$$=\int \ddot{x}t+\dot{x_0} dt$$
$$=\dot{x_0}t+\frac{\ddot{x}}{2}t^2+c$$

Mentor
Ohh! Thanks... got it...

$$\dot{x} (t)=\int \ddot{x} t dt=\ddot{x}t+c$$
$$\dot{x}(t)=\ddot{x}t+\dot{x}(0)$$
$$x(t)=\int \dot{x} (t) dt$$
$$=\int \ddot{x}t+\dot{x_0} dt$$
$$=\dot{x_0}t+\frac{\ddot{x}}{2}t^2+c$$
OK, much better!

And it looks like your book assumed a constant acceleration (because of a constant force) equal to F0/m.

Istiak
Istiak
OK, much better!

And it looks like your book assumed a constant acceleration (because of a constant force) equal to F0/m.
Hum..!

Fred Wright
I don't understand the confusion. We start with a constant force along the x-axis, ##F_0##. Newton's law says,
$$F=ma=m\frac{d^2x}{dt^2}=F_0$$
$$m\frac{d^2x}{dt^2}=m\frac{d\dot x }{dt}$$
$$\frac{d\dot x }{dt}=\frac{F_0}{m}$$
Assuming ##t_0=0## we integrate
$$\int_{\dot {x}_0}^{\dot x}d\dot {x}'=\int_0^t\frac{F_0}{m}dt'$$
with the primes indicating dummy variables of integration.
$$\dot x - \dot {x}_0=\frac{F_0}{m}t$$
$$\dot {x}_0=v_0$$
$$\dot x=\frac{dx}{dt}$$
$$\frac{dx}{dt}= v_0+\frac{F_0}{m}t$$
Integrate again
$$\int_{x_0}^x dx'=\int_{0}^t (v_0+\frac{F_0}{m}t')dt'$$
$$x=x_0 + v_0t+ \frac{F_0t^2}{2m}$$
Does this help or am I missing something?