# A Physics type of integral

1. Jul 17, 2003

### Suicidal

In my calculus course I was tough to solve differential equations by separation and then integrating
For instance
dv/dt = a
dv = a*dt
v= a*t + c (c is a constant of integration)
then if I was given an initial condition such as v(0)=Vo I would substitute into my general solution
and get
Vo=a*0+c=c
V=a*t+Vo
Now that I started learning physics I keep coming across this funny way of solving differential equations:
v
[inte] dv=
Vo
t
[inte] a*dt
0
Why does this second method work. I realize that it is probably somehow equivalent to the first I just don’t see it. I would really like to know how and why this second method works.

2. Jul 17, 2003

### Kalimaa23

What you are doing is basically applying the chain rule
You have an original fuction f(v). It turns out that v itself is a fuction of t, say v(t).
The differential gives
dv = v'(t)dt
or in Leibnitz notion
dv = (dv/dt) dt. You can see this be saying the dt's cancel out. This is of course wrong, as any mathematician will tell you, but in practice it makes working with differential easier.

Now all you do is apply the chain rule to the given integral so that

v(t)
[inte]dv =
v(0)

t
[inte](dv/dt)dt with then saying physics-wise that dv/dt=a
0

Say when you replace the funtion of v the function of t, you adjust the boundries of the integral by putting the argument of the function there instead of the it's value. So generally putting a instead of f(a).

Consult any freshman calculus text for a brief proof of this if you don't believe me (and you shouldn't )

3. Jul 17, 2003

### Suicidal

You answer doesn’t really answer my questing. I was basically asking why the method works, so I guess I was asking for a proof.

I have taken Calculus BC in high school which is equivalent to Calculus I & II and I have a book called Thomas’ Calculus that goes through all the way to Calculus III. I have never seen this method of solving differential equations used outside my physics book.

My calc book doesn’t include any proof for this method because it doesn’t use it.

Is anyone familiar with this “brief proof”?

4. Jul 17, 2003

### Hurkyl

Staff Emeritus
This really is the same thing as the chain rule.

In general, seperation of variables means that you can write your differential equation as:

f(y) dy = g(t) dt

Which, at the calculus level, really means:

f(y(t)) (dy(t) / dt) = g(t)

Consider the integral

&int;y(a)..y(b) f(y) dy

(where I've changed notation for an integral slightly due to the fact superscripts don't tend to look nice in this context)

Since y is a function of t, we may apply the chain rule to get

&int;a..b f(y(t)) (dy(t) / dt) dt = &int;a..b g(t) dt

5. Jul 18, 2003

### Suicidal

I think I see it now.

Thank you both.
And Dimitri Terryn I am sorry for saying that you weren't answering my question. I simply misunderstood what you were saying. After reading Hurkyl's answer I realized that you were both saying the same thing.

6. Jul 18, 2003

### Kalimaa23

No problem. I've only just finished my first year math/physics so I haven't got alot of experience explaining these things to other people. This is one of the reasons I hang around here, trying to explain something invariably helps my own understanding.