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?

 

[fresh_42]

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

 

[Doc Al]

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

 

[Istiak]

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

$$\dot{x}$$ is velocity... It's obviously constant [as far as I know]( https://math.codidact.com/posts/283445/283449#answer-283449 )

 

[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

 

[Doc Al]

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?

 

[Doc Al]

What my book didn't do?

It looks to me as if the problem assumes a constant acceleration, not a constant velocity.

 

[fresh_42]

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]

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 $$x(t)=x(0)+vt+\frac{at^2}{2}$$

 

[Doc Al]

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.

 

[Istiak]

You assume that the velocity is constant, but your book does not.

How? If I integrate velocity than, I get position not acceleration...

 

[Doc Al]

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.

 

[fresh_42]

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 $c$.

 

[Doc Al]

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...

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

 

[Doc Al]

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]

OK, much better!

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

Hum..!

 

[fresh_42]

You can find two examples of O(rdinary)D(ifferential)E(quations) here:

Question #7:
https://www.physicsforums.com/threads/basic-math-challenge-june-2018.948674/

Question #5:
https://www.physicsforums.com/threads/math-challenge-january-2021.997970/

 

[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?

 

[haruspex]

We start with a constant force along the x-axis

How do you know that? I don't see any links in this thread to the original question.