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How do you find the general solution of differential equations?

  1. Nov 2, 2004 #1
    I really don't understand the concept of differential equations! I'm confused about how you would go about solving them.

    if (x^2).dy/dx = y^2,

    integral(y^2)dy = integral(x^2)

    1/3y^3 = 1/3x^3 + A

    ^ What have I done wrong here?

    I don't get the whole thing where you separate the variables etc.

    Am I right in saying that integral(dy/dx)dx = y?

    Moderators, sorry I posted this twice. It was an accident.
    Please can someone help me understand all of this? I understand basic differentiation and integration.
  2. jcsd
  3. Nov 2, 2004 #2
    Just put all variables of one sort at one side : x²dy/dx = y². This means that (1/y²)dy = (1/x²)dx and just integrate --> -(1/y) + C = -(1/x) + C'

  4. Nov 2, 2004 #3
    [tex]x^{2}\frac{dy}{dx} = y^{2}[/tex]
    If you rearrange that you'd see that it equals:
    [tex]\frac{1}{y^2}dy = \frac{1}{x^2}dx[/tex]
    And not:
    [tex]y^2 dy = x^2 dx[/tex]

    To solve the equation, simply take the integral of both sides.
  5. Nov 2, 2004 #4
    So you substitute y for x and vice versa, take the reciprocal of both variables and then integrate?

    I don't understand how you got from the first step to the second step (and then to the third)!
  6. Nov 2, 2004 #5
    well the integral of (1/x²) is -(1/x), that's all...

    use the rule integral of x^n = (x^(n+1)/n+1) and n = -2 here...
  7. Nov 2, 2004 #6
    I understand that, but I don't get why you substituted x^2 for 1/y^2 and y^2 for 1/x^2.

    It's not integrating of individual terms I'm confused about...
  8. Nov 2, 2004 #7
    It's simple manipulation:

    [tex]a\;\frac{b}{c}=d \Rightarrow \frac{1}{d}\;b=\frac{1}{a}\;c[/tex]
  9. Nov 2, 2004 #8
    I thought that dy/dx is not a fraction (and therefore cannot be rearranged)?
    Last edited: Nov 2, 2004
  10. Nov 2, 2004 #9

    matt grime

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    dy/dx is indeed not a fraction, but there are *some* circumstances where it can be treated as one, and this is one of them. We're just simply taking a practical approach to how to solve the equation, that is omitting some steps that always occur in the same way, to clear up the presentation.

    What it is is a short hand wayo of saying that if:

    f(y)dy/dx = g(x)

    after rearranging, then

    int f dy = int g dx
  11. Nov 2, 2004 #10
    What do f dy and g dx mean? Don't you mean int f(y)dy = int f(x)dx?

    I don't get what f and g on their own mean.
  12. Nov 2, 2004 #11

    matt grime

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    yes f on its own means f(y), but on the right hand side it is int g(x)dx

    normally, where there is no room for confusion, it is acceptable, for example, to drop the letter t from h(t) and refer to the function simply as h.
  13. Nov 2, 2004 #12
    You can treat the dy/dx like a fraction because it essentially is one. It is the ratio of two differential operators. So when you multiply both sides by the differential operator dx, you get cancellation on one side and a dx on the other.
  14. Nov 2, 2004 #13
    ^ Thank you. That's what I thought!
  15. Nov 2, 2004 #14

    matt grime

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    dx isn't a differential operator (it doesn't take a function and yield another function, but you don't want to get too technical about that) they are referred to as infinitesimals occasionally.
  16. Nov 2, 2004 #15
    It’s not? What is it classified as then? It’s some kind of operator right?
  17. Nov 2, 2004 #16
    just a differential arn't they ?

    depending on whether the indeterminant is a dependent or independent variable

    [tex]dx = \Delta x[/tex]
    [tex]dy = f'(x) dx[/tex]

  18. Nov 3, 2004 #17

    matt grime

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    Hope this doesn't confuse anyone unnecessarily, but dx is the dual vector to the vector [tex]\partial_x[/tex], and is called a 1-form. It is an element of the cotangent bundle, but it is not a differential operator. (Most things can be thought of as "operators" I suppose, it's one of those frequently used ambiguous labels). A differential operator looks something like:

    [tex] \mathcal{L}= \partial_{xy} - \partial_{x} +x^2\partial{_y}[/tex]
  19. Nov 3, 2004 #18


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    No, its a "differential", not a differential "operator".

    I would be very reluctant to tell anyone "You can treat the dy/dx like a fraction because it essentially is one."

    In standard calculus, dy/dx is defined as the limit of the fraction Δy/Δx, NOTas the fraction "dy" over "dx". One can then show that, generally, one can treat dy/dx as if it were a fraction by going back before the limit, using the fact that Δy/Δx is a fraction and then going forward again. Typically, one then defines the "differentials" dy and dx separately in a purely symbolic manner.

    Yes, it is possible to define the derivative in terms of "infinitesmals", dx and dy, from the start ("non-standard calculus") but to define "infinitesmals" rigorously requires some very deep logical gymnastics!
  20. Nov 3, 2004 #19

    matt grime

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    Another caveat, and a demonstrable one, is that although dy/dx behaves as a fraction, second and higher derivatives do not (nor do partial derivatives)

    everyone knows that (dy/dx)^{-1}= dx/dy but it is certainly not true that

    (d^2y/dx^2)^{-1} = d^2x/dy^2

    and it is quite instructive to actually try and work out the proper relationship.
  21. Nov 3, 2004 #20


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    In fact, that's one of the reasons the superscript numbers are place in different positions in the "numerator and denominator": in order that
    1/(d2y/dx2)= dx2/d2y makes no sense!
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