How do the d terms in differential equations work?

[Chrisistaken]

How do the "d" terms in differential equations work?

Hi,

I was hoping someone could explain how the "d" terms in differential equations work? For example,

d²y/dx² = 4x³ +1

To solve I have been rearranging to get,

d²y = (4x³ +1)dx²

and then doing a double integral of each side.

I sort of assume that the double integral of d²y is equal to y rather than y²/2 but I don't really understand why this is the case.

Can someone please explain this to me and also why the powers are applied differently to the numerator and the denominator?

Regards,

Chris

 

[charlesmanima]

The integral of d²y is dy. Once again integrating you will get y. The way you are rewriting the equations can make confusion. To avoid this confusion rewrite d²x/dx² as d/dx(dy/dx). Now you can integrate two times with respect to x.

 

[chiro]

The integral of d²y is dy. Once again integrating you will get y. The way you are rewriting the equations can make confusion. To avoid this confusion rewrite d²x/dx² as d/dx(dy/dx). Now you can integrate two times with respect to x.

Umm isn't d/dx(dy/dx) = d^2(y)/dx^2?

 

[Hurkyl]

Hi,

I was hoping someone could explain how the "d" terms in differential equations work? For example,

You're grouping symbols wrong. Roughly speaking everything in

$$\frac{d}{dx}$$​

goes together into one symbol. (though, I suppose, x is a slot for inserting an appropriate variable symbol)


AFAIK, the assortment of coincidences that allow us, when there is only one independent variable, to get sensible results by viewing $dy/dx$ as if it was the quotient of $dy$ and $dx$ don't work very well when applied to second derivatives.

You're better off working one derivative at a time -- e.g. worry first about

$$d \left( \frac{dy}{dx} \right) = (4x^3 + 1) dx$$​

It might help to introduce a new variable

$$z := \frac{dy}{dx}$$​

so the previous equation becomes

$$dz = (4x^3 + 1) dx$$​

 

[HallsofIvy]

Hi,

I was hoping someone could explain how the "d" terms in differential equations work? For example,

d²y/dx² = 4x³ +1

To solve I have been rearranging to get,

d²y = (4x³ +1)dx²

No, this is not valid. The first derivative is defined as a limit of a fraction and so can be "treated" like a fraction (to show that any property of a fraction is true for a derivative, go back before the limit, use that property of the difference quotient, then take the limit). The differentials, dy and dx, are defined from the derivative to make use of that.

But higher order derivatives are not defined like that and cannot be treated as fractions. In particular, saying that $d^2y/dx^2= f(x,y)$ does NOT lead to "$d^2y= f(x,y)dx^2$.

and then doing a double integral of each side.

No, that's not the way double integrals work. You have to have two different variables, not $\int d^2y$

I sort of assume that the double integral of d²y is equal to y rather than y²/2 but I don't really understand why this is the case.

Can someone please explain this to me and also why the powers are applied differently to the numerator and the denominator?

In order to remind you that higher order derivatives are NOT just fractions!

Regards,

Chris

 

[Chrisistaken]

Thanks all for the quick replies. I think Charlesmanima pretty much summed up what I was after.

Just to clarify though, are the d/dx, dx and dy terms, just symbols that cannot be manipulated by the usual mathematical methods (excluding convenient coincidences)?

Regards

Chris.

 

[pmsrw3]

Can someone please explain this to me and also why the powers are applied differently to the numerator and the denominator?

In order to remind you that higher order derivatives are NOT just fractions!

Also (or maybe as a specific example of what HallsofIvy is saying), it makes the units work out right. For instance, $\frac{dx}{dt}$ is velocity, which has units of distance/time, like $\frac{x}{t}$. $\frac{d^2x}{dt^2}$ is acceleration, which has units of distance/time², like $\frac{x}{t^2}$.

 

[Chrisistaken]

So for a fail safe method of working with derivatives, you would need to go back to the limit?