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Understanding Separable DEs

  1. Feb 25, 2013 #1
    ∫ y' dx = ∫ dy
    ∫ yy' dx = ∫ y dy

    I can't clearly visualize how this is working, and it's messing with me when I work with equations that flip between displacement, velocity, and acceleration. Could someone link me to a proof, or maybe explain it a little for me?

    Thanks!
     
  2. jcsd
  3. Feb 25, 2013 #2

    LCKurtz

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    Remember that if y is a function of x, the differential dy is defined to be ##\frac{dy}{dx}dx## and that is the substitution being used. Now, the left integral is just the antiderivative of y', which gives y as the answer, ignoring the constant. The integral on the right is ##\int 1\, dy## which also gives ##y## as the answer. So the formal substitution of dy = y'dx is OK to use.

    For the second it is the same substitution dy = y'dx. Again, if you work the left side it is just the antiderivative of yy' which is ##\frac 1 2 y^2##, as you can verify by chain rule differentiation. And that is the same answer you get by working the right side as if y were an independent variable. So, again, the method works. Is is really just an application of the chain rule in reverse.
     
  4. Feb 25, 2013 #3

    LCKurtz

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    After I posted my above reply, I noticed the title was "understanding separable equations". So let me add a little more explanation. When you successfully separate variables in a DE of the form y'=f(x,y) you arrive at a form like this:$$
    n(y)\frac{dy}{dx} = m(x)$$
    At this point we "separate the variables" by the magical multiplying both sides by dx:
    $$
    n(y)dy=m(x)dx$$If that doesn't already bother you, then what about the fact that you now integrate one side with respect to x and the other with respect to y? Ordinarily, after all, you must do the same thing to both sides of an equation. If you go ahead and do that, you get ##N(y) = M(x)+C## where ##M## and ##N## are antiderivatives of ##m## and ##n##, and you claim that is the solution.

    But is it really? The way to find out is to see if it satisfies the original DE which we have written in the form$$
    n(y)\frac{dy}{dx} = m(x)$$
    Let's differentiate both sides of our claimed solution ##N(y)=M(x)+C## with respect to ##x##:$$
    \frac d {dx}N(y) = \frac d {dy}N(y)\frac {dy}{dx}=\frac d {dx}M(x)+0$$
    by the chain rule. This last equality can be written$$
    N'(y) \frac {dy}{dx}=M'(x)$$ $$
    n(y)\frac {dy}{dx} = m(x)$$so it does satisfy the equation.

    Separation of variables is thus justified and it works because it is a shortcut way of reversing the chain rule.
     
    Last edited: Feb 25, 2013
  5. Feb 25, 2013 #4
    Thank you for your post, I tried to write a proof to see if I understand it. Is this correct logic?

    By definition of a derivative
    Rate of change ≈ [itex]\frac{Δy}{Δx}[/itex]
    Rate of change = [itex]lim_{Δx→0}[/itex] [itex]\frac{f(x + Δx) - f(x)}{Δx}[/itex]

    Integrate

    [itex]lim_{Δx→0}[/itex][itex]\sum \left[lim_{Δx→0}\frac{f(x + Δx) - f(x)}{Δx}\right][/itex]*Δx
    [itex]lim_{Δx→0}[/itex][itex]\sum \left[\frac{f(x + Δx) - f(x)}{Δx}\right][/itex]*Δx
    [itex]lim_{Δx→0}[/itex][itex]\sum \frac{f(x + Δx) - f(x)}{1}[/itex]
    [itex]lim_{Δx→0}[/itex][itex]\sum Δy[/itex]

    As Δx→0, Δy→0

    [itex]lim_{Δy→0}[/itex][itex]\sum Δy[/itex] = [itex]\int[/itex] dy


    Is this right?
     
  6. Feb 25, 2013 #5

    LCKurtz

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    No, that doesn't do anything for me. I don't even know what it purports to prove. Did you read my second post above (#3)?
     
  7. Feb 25, 2013 #6
    Was trying to show that ∫ y' dx = ∫ dy by using definitions of derivatives and integrals. Did I make a mistake in there, or is it just confusing? I did read yours! "This works because the assumed answer works." I follow perfectly, just curious if this is also another way to show the equality.
     
  8. Feb 25, 2013 #7

    LCKurtz

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    That would be a complicated way to do it if it did work. But as it is, it doesn't make much sense. For example, in the two steps I have included above, one of the limits mysteriously disappeared. Perhaps you are confusing the derivative with the difference quotient. But there are enough other problems with that argument that if I were you, I would forget about trying to prove it that way.
     
  9. Feb 25, 2013 #8
    That's me not knowing what I'm doing! I was assuming that if Δx→0 for both parts of the equation, you can simply pull the limits together. It seems like an intuitive step, but I don't know if it's actually valid.
     
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