- #1

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Could you, for instance, replace [itex]\frac{dy}{dx}[/itex] with [itex]y'(x)[/itex] in a differential equation and work it out?

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- Thread starter jaydnul
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- #1

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Could you, for instance, replace [itex]\frac{dy}{dx}[/itex] with [itex]y'(x)[/itex] in a differential equation and work it out?

- #2

Simon Bridge

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You can treat the Liebnitz notation as a fraction because that's the way Liebnitz worked out the notation. In that sense it is not "just" a notation.

You can use the primed notation - many do - it's just harder to make the relations.

You can solve, for eg. ##y'(x)=a## by integrating both sides wrt x: $$\int y'\cdot dx = a\int dx \Rightarrow y=ax+c$$

But what about ##y'(x)=ay## ?

You can use the primed notation - many do - it's just harder to make the relations.

You can solve, for eg. ##y'(x)=a## by integrating both sides wrt x: $$\int y'\cdot dx = a\int dx \Rightarrow y=ax+c$$

But what about ##y'(x)=ay## ?

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- #3

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[tex]∫ydx=e^x+c[/tex]

So how would I solve this using prime notation:

[tex]y'(x)=y[/tex]

- #4

Simon Bridge

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I had hoped you'd do something like this: $$\frac{dy}{dx}=ay\\ \Rightarrow \int \frac{dy}{y}=a \int dx \\ \Rightarrow \ln |y|=ax+c \\ \Rightarrow y=e^{ax+c}=e^ce^{ax}=Ce^{ax}$$Using [itex]\frac{dy}{dx}[/itex] notation, I get:

[tex]∫ydx=e^x+c[/tex]...

That's my point: there's no systematic way like there is with Leibnitz - you'd have to recognize the form of the equation and apply a rule or use guesswork.So how would I solve this using prime notation:

[tex]y'(x)=y[/tex]

You'd notice it is of form ##\sum a_n y^{(n)}=0## and

- #5

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So how did newton do his differential equations?

Btw Simon, I really appreciate the help!

Btw Simon, I really appreciate the help!

- #6

Simon Bridge

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Presumably if he had to use infinitesimal differentials he'd treat them as an inverse integration... using his experience of integration to guide his guesses. Substitute in as above.

Notice: ##\int y'(x)dx = y## right? So integrating both sides of the example he'd have got ##y=\int y(x)dx## and he'd look for equations which have that property. What integrates to give itself? It's just the same as when you do ##\small 3x=12## - you are asking what, multiplied by 3, gives you 12. You do it by experience and look-up tables (which you have memorized).

Note: The number one main method for solving DEs remains the guesswork method.

We just know a lot of tricks for figuring out a good guess.

- #7

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Ya that makes sense because the only function of x that satisfies [itex]y=y'[/itex] and [itex]y=∫y[/itex] would be [itex]e^x[/itex].

I can see how that would get complicated haha.

Thanks again Simon!

- #8

Simon Bridge

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Newtons notation is compact though, and easier to type, so you'll still see it a lot for written notes.

No worries - see that "thanks" button?

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