Why does integrating both sides of a differential equation work in physics?

In summary, the conversation discusses the different methods for solving differential equations, particularly the use of separation of variables and the chain rule. The second method involves transforming the differential equation into an integral and applying the chain rule to solve it. This method is equivalent to the first method but may be easier to work with in practice. The conversation also mentions the use of a "brief proof" for this method, which can be found in most freshman calculus textbooks.
  • #1
Suicidal
22
1
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.
 
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  • #2
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 :smile: )
 
  • #3
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
This really is the same thing as the chain rule.


In general, separation 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

∫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

∫a..b f(y(t)) (dy(t) / dt) dt = ∫a..b g(t) dt
 
  • #5
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
No problem. I've only just finished my first year math/physics so I haven't got a lot 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.
 

1. What is a Physics type of integral?

A Physics type of integral is a mathematical calculation used in the field of physics to determine quantities such as displacement, velocity, acceleration, and work. It involves finding the area under a curve on a graph and is often used in solving problems related to motion and forces.

2. How is a Physics type of integral different from a regular integral?

A Physics type of integral is different from a regular integral in that it is specifically used in the context of physics problems. It involves applying principles of physics, such as Newton's laws of motion, to set up the integral and solve for a physical quantity. A regular integral, on the other hand, can be used in various fields of mathematics.

3. What are some common applications of Physics type of integrals?

Physics type of integrals are commonly used in problems related to motion, such as calculating the displacement, velocity, or acceleration of an object. They are also used in calculating work done by a force, finding the center of mass of an object, and determining the total energy of a system.

4. Can Physics type of integrals be solved without calculus?

No, Physics type of integrals require knowledge of calculus in order to be solved. They involve finding the antiderivative of a function and evaluating it at specified limits. Without calculus, it would not be possible to solve for the desired physical quantity.

5. Are there any tips for solving Physics type of integrals more efficiently?

Some tips for solving Physics type of integrals more efficiently include identifying the physical principles involved, setting up the integral correctly, choosing appropriate limits of integration, and using integration techniques such as substitution or integration by parts. It is also helpful to practice solving different types of integrals to become more familiar with the process.

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