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Homework Help: Help with Differential Equations (First order)

  1. Nov 26, 2014 #1
    1. The problem statement, all variables and given/known data
    Solve the given the two equations:

    [tex] xdy + ydx = ydy[/tex]


    [tex](y^2 + 1)dx +(2xy + 1)dy = 0 [/tex]

    2. Relevant equations


    3. The attempt at a solution

    For the first, I can see that solving this with respect to [tex]dy/dx[/tex]might be a bit tricky.

    However, if I solve it for [tex]dx/dy[/tex], things seem to be a bit easier.

    [tex] xdy + ydx = ydy [/tex]

    [tex] ydx = ydy - xdy[/tex]

    [tex] y\frac{dx}{dy} + x = y [/tex]

    Now, that looks a bit familiar. Did someone say "product rule"?

    [tex] \int{yx\frac{dx}{dy}} = \int{ydy} [/tex]

    [tex] yx = \frac{y^2}{2} [/tex]

    [tex] x = \frac{y}{2} [/tex]

    [tex] y = 2x + C [/tex]

    Now, this seems a bit too easy to me. Is that really it?

    As to the second...

    [tex] (y^2 + 1)dx + (2xy+1) dy = 0 [/tex]

    [tex] y^2 dx + 2xy dy = -dy - dx [/tex]

    [tex] y^2 + 2xy\frac{dy}{dx} = -\frac{dy}{dx} - 1 [/tex]

    Once again, it seems like that product rule comes up quite a bit...

    [tex] \int{y^2 x\frac{dy}{dx}} = -\int{\frac{dy}{dx} - 1} [/tex]

    At this point, I got a sinking feeling that something was horribly wrong with my work. I checked my answers on wolfram alpha, got a bunch of gibberish that made no sense, as well as answers that in no way resembled my own work. Could anyone verify if I'm taking the correct steps?

    Last edited: Nov 26, 2014
  2. jcsd
  3. Nov 26, 2014 #2


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    Getting to the last equation was an error. That should be:
    [itex]y \frac{dx}{dy} + x = y[/itex]
  4. Nov 26, 2014 #3
    Doh! Fixed that, sorry.
  5. Nov 27, 2014 #4


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    Homework Helper

    You miss an integration constant here.
  6. Nov 27, 2014 #5


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    Homework Helper

    The last line is wrong. As ## y^2 + 2xy\frac{dy}{dx}=\frac{d(y^2x)}{dx}##, it should be
    [tex] \int{\frac{d(y^2 x)}{dx}dx} = -\int{(\frac{dy}{dx} + 1)dx} [/tex]
  7. Nov 28, 2014 #6


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    Personally, I dislike that particular method. It works- you can simply integrate- because these are exact equations.

    The first, xdy+ ydx= ydy, is the same as ydx+ (x- y)dy= 0. Saying it is exact means there exist a function, F(x, y), such that
    [tex]dF= F_x dx+ F_y dy= ydx+ (x- y)dy[/tex].
    From [tex]F_x= y[/tex] we have, immediately, that F= xy+ g(y). From that, [tex]F_y= x+ g'(y)= x- y[/tex] so that [tex]g'(y)= -y[/tex]. Integrate that to find g.
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