Solving differential equations with different differential notation?

  • Thread starter jaydnul
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  • #1
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Since [itex]\frac{dy}{dx}[/itex] is just considered notation, how can we treat it as an actual fraction when soliving differential equations?

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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  • #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## ?
 
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  • #3
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Using [itex]\frac{dy}{dx}[/itex] notation, I get:
[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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Using [itex]\frac{dy}{dx}[/itex] notation, I get:
[tex]∫ydx=e^x+c[/tex]...
I had hoped you'd do something like this: $$\frac{dy}{dx}=ay\\ \Rightarrow \int \frac{dy}{y}=a \int dx \\ \Rightarrow \ln |y|=ax+c \\ \Rightarrow y=e^{ax+c}=e^ce^{ax}=Ce^{ax}$$
So how would I solve this using prime notation:
[tex]y'(x)=y[/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.

You'd notice it is of form ##\sum a_n y^{(n)}=0## and guess ##y=e^{\lambda x}## and substitute the guess into the equation to find lambda.
 
  • #5
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So how did newton do his differential equations?

Btw Simon, I really appreciate the help!
 
  • #6
Simon Bridge
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Newton's contribution was primarily in integration, which he could verify geometrically. If you look through Principia, for eg. you'll see he didn't use calculus for any of his proofs.

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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Oh I see.

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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Oh sure - DEs get arbitrarily complicated even at 1st order.
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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