You are Here: Home >< Maths

# Question about differential equations? watch

1. During my mechanics course the professor had an equation like:

he then split the fraction up like this and integrated like so to find a formula for . (Can't remember if this was the expression but it is irrelevant.)

I was wondering how you can do such a thing (what is really going on and what justifications have to be made in order to do such a thing)?

I remember treating the derivative as a fraction before at A-Level and not knowing what I was actually doing so I have decided to try and fix that.

I have know the formal definitions of derivatives, riemann integrals and limits (and just now seeing the fundamental theorem of calculus.)

I haven't taken any modules in differential equations and I'm not really sure if it is beyond my scope at the moment.

Could anyone explain or give me a rough idea of what is actually going on or point me in the direction of some resources I can look at?

Cheers.
2. I think this is called the variable separable method of integration. (I learnt about it a while ago). Pretty useful in C4 and FP2-3. Sorry I can't help more!

Posted from TSR Mobile
3. (Original post by alex2100x)
Could anyone explain or give me a rough idea of what is actually going on or point me in the direction of some resources I can look at?

Cheers.
From what I can tell, he's simply integrating both sides of the equation with respect to t. Think back to implicit differentiation: because both sides are equal, their differentials must be equal, too. This is the same concept, but with integrals instead.
The integral of with respect to t is just .
As for the right hand side of the equation, it's pretty much the same story, except the dt essentially cancels out.

Look at that bit as a fraction, and consider what's actually going on in terms of gradients/areas. Put simply, the fraction's got dt on the bottom and by integrating, you're multiplying by dt, so those terms cancel out because, obviously, dt/dt is just 1.

That's probably the simplest way of looking at it.
4. It can be justified by considering that:

since we have just differentiated then integrated the result of that differentiation - these are inverse operations (up to the constant ) so we end up where we started. Also:

so these expressions are equivalent
5. Ok thanks I guess it relies on FTOC to justify that differentiating and integrating are inverses of each other.

cheers.

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: January 27, 2015
Today on TSR

### Tuition fees under review

Would you pay less for a humanities degree?

### Can I get A*s if I start studying now?

Discussions on TSR

• Latest
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE