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What is a first order differential equation?

In the C4 book it says 'you will consider only first order differential equations, which involves first derivatives only'.

What is a first order differential equation and how does it differ from a second order differential equation (if that's what it's called).

I also often see people correcting others incorrect usage of differential and derivative, what's the difference?
(edited 7 years ago)
All you differential equation needs https://en.wikipedia.org/wiki/Differential_equation
Reply 2
Avatar for TeeEm
TeeEm
19
too late as always
In a first order differential equation you are looking for functions which satisfy certain conditions relating to the function and it's first derivative.
With a second order differential equation you are looking for functions which satisfy certain conditions relating to the function and it's first and second derivatives.
(edited 8 years ago)
Reply 4
Avatar for Zacken
Zacken
22
Original post by NotNotBatman
In the C4 book it says 'you will consider only first order differential equations, which involves first derivatives only'.

What is a between a first order differential equation and how does it differ from a second order differential equation (if that's what it's called).

I also often see people correcting others incorrect usage of differential and derivative, what's the difference?


A differential equation is an equation that is made up functions and its derivatives. A first order differential equation is one that contains up to first order derivatives. A second order differential equation is one that contains up to second order derivatives.

A differential equation is made up of derivatives. But a derivative is not the same as a differential.
Original post by TeeEm
too late as always


Could be better than being too early... :u:
Reply 6
Avatar for TeeEm
TeeEm
19
Original post by EricPiphany
Could be better than being too early... :u:


not too bad ...
many more people later than !
Original post by Zacken
A differential equation is an equation that is made up functions and its derivatives. A first order differential equation is one that contains up to first order derivatives. A second order differential equation is one that contains up to second order derivatives.

A differential equation is made up of derivatives. But a derivative is not the same as a differential.


So if f is a differential equation it can be differentiated to nth order and an nth order differential equation is one that can be differentiated n times, resulting in an nth order derivative. Additionally, differentiating is the process that leads to the derivative from a differential equation which contains the derivative as a composite function.

This is what I'm thinking given your explanation, is it correct?
Reply 8
Avatar for Zacken
Zacken
22
Original post by NotNotBatman
So if f is a differential equation it can be differentiated to nth order and an nth order differential equation is one that can be differentiated n times, resulting in an nth order derivative. Additionally, differentiating is the process that leads to the derivative from a differential equation which contains the derivative as a composite function.

This is what I'm thinking given your explanation, is it correct?


I'm not sure why you're differentiating differential equations. A first order differential equation is something like dy/dx + x = y. You solve it for y, you don't diffentiate it. Notice that the highest order derivative in that equation is dy/dx = d^1y / dx^1 which is first order.
Original post by Zacken
I'm not sure why you're differentiating differential equations. A first order differential equation is something like dy/dx + x = y. You solve it for y, you don't diffentiate it. Notice that the highest order derivative in that equation is dy/dx = d^1y / dx^1 which is first order.


Oh, I did this a few months ago, separating the variables, I was confused because I haven't done it in C4 classes yet.

I have another question.
Why is it that dy/dx is treated as a fraction; in seperating the variables it's sometimes explained as 'multiplying both sides by dx, you're not really, but it works like this' and in the chain rule sometimes it is explained using leibniz notation as dy/du . du/dx and the du cancels.

I've seen the workings from first principles, but why can it be worked out treating it as a fraction if it isn't?
Reply 10
Avatar for Zacken
Zacken
22
Original post by NotNotBatman
Oh, I did this a few months ago, separating the variables, I was confused because I haven't done it in C4 classes yet.

I have another question.
Why is it that dy/dx is treated as a fraction; in seperating the variables it's sometimes explained as 'multiplying both sides by dx, you're not really, but it works like this' and in the chain rule sometimes it is explained using leibniz notation as dy/du . du/dx and the du cancels.

I've seen the workings from first principles, but why can it be worked out treating it as a fraction if it isn't?


It just so happens that if you pretend that they're fractions, it works out. Quite convenient. I mean, afterall dy/dx is kiiiiind of a fraction. It can be thought of as the infinitesimal change in y over the infinitesimal change in x, but that's not technically true. To be honest, I don't have a very indepth answer, these things stem from non-standard analysis and I've yet to scratch much of that field. Perhaps @atsruser might be able to shed some light with his usual (great!) explanations.
Original post by NotNotBatman
Oh, I did this a few months ago, separating the variables, I was confused because I haven't done it in C4 classes yet.

I have another question.
Why is it that dy/dx is treated as a fraction; in seperating the variables it's sometimes explained as 'multiplying both sides by dx, you're not really, but it works like this' and in the chain rule sometimes it is explained using leibniz notation as dy/du . du/dx and the du cancels.

I've seen the workings from first principles, but why can it be worked out treating it as a fraction if it isn't?


Maybe because it works?

...and it works in lots of cases most probably because δy≈dydxδx\displaystyle \delta y \approx \frac{dy}{dx}\delta x, so it looks like you can cancel the fractions.
Also dydx=dydtdtdx\displaystyle \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}
(edited 8 years ago)
Reply 12
Avatar for Zacken
Zacken
22
Original post by DuckIsBackBRAP
How have you been you fat ugly paki muddafuka


I've been good.
Original post by Zacken
I've been good.


I n - 1ed your mum last night

n being your followers count

how is she
Original post by Zacken
It just so happens that if you pretend that they're fractions, it works out. Quite convenient. I mean, afterall dy/dx is kiiiiind of a fraction. It can be thought of as the infinitesimal change in y over the infinitesimal change in x, but that's not technically true. To be honest, I don't have a very indepth answer, these things stem from non-standard analysis and I've yet to scratch much of that field. Perhaps @atsruser might be able to shed some light with his usual (great!) explanations.


It works due to the chain rule. Suppose:

∫f(u) dx=F(u)+c⇒dFdu=f(u)\int f(u) \ dx = F(u) + c \Rightarrow \frac{dF}{du} = f(u)

Then if u=g(x)u=g(x) then

F(u)=F(g(x))⇒dFdx=dFdududx=f(u)dudxF(u) = F(g(x)) \Rightarrow \frac{dF}{dx} = \frac{dF}{du} \frac{du}{dx} = f(u) \frac{du}{dx}

If we now integrate both sides w.r.t xx we have:

∫f(x) dx=F(x)+c=∫dFdx dx=∫f(u)dudx dx\int f(x) \ dx = F(x) + c = \int \frac{dF}{dx} \ dx = \int f(u) \frac{du}{dx} \ dx

So if we have say dydx=yx⇒1ydydx=1x\frac{dy}{dx} = \frac{y}{x} \Rightarrow \frac{1}{y}\frac{dy}{dx} = \frac{1}{x} we can write either:

∫1ydydxdx=∫1x dx+c\int \frac{1}{y} \frac{dy}{dx} dx = \int \frac{1}{x} \ dx + c

or:

∫1y dy=∫1x dx+c\int \frac{1}{y} \ dy = \int \frac{1}{x} \ dx + c

interchangeably.

If we restrict ourselves to polynomials, we can also show this using IBP since:

∫xndxdu du=xnx−∫xd(xn)du du+c=xn+1−∫xd(xn)dxdxdu du+c\int x^n \frac{dx}{du} \ du = x^n x - \int x \frac{d (x^n)}{du} \ du + c = x^{n+1} - \int x \frac{d (x^n)}{dx}\frac{dx}{du} \ du +c

=xn+1−n∫xndxdu du+c = x^{n+1} - n \int x^n \frac{dx}{du} \ du +c

So (n+1)∫xndxdu du=xn+1+c⇒∫xndxdu du=xn+1n+1+c=∫xn dx(n+1) \int x^n \frac{dx}{du} \ du = x^{n+1} +c \Rightarrow \int x^n \frac{dx}{du} \ du = \frac{x^{n+1}}{n+1} + c = \int x^n \ dx

but I don't think that argument can be generalised.

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