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Is this the correct solution to this DiffEq?

  1. Dec 29, 2013 #1
    $$ xy' = y + xy^2 $$
    $$ xy' - y = xy^2 $$
    $$ y' + (-1/x)y = y^2 $$
    $$ e^{∫(-1/x)dx} = 1/x $$
    $$ (1/x)[y' + (-1/x)y = y^2] $$
    $$ (1/x)y' - (1/x^2)y = (1/x)y^2 $$
    $$ ∫((1/x)y)' = ∫((1/x)y^2) dx $$
    $$ (1/x)y = (y^2)ln|x| + C $$
    $$ y = (xy^2)ln|x|+Cx $$

    Wolfram shows me another solution...
    $$y = \frac{-2x}{C+x^2}$$
    Are both correct?
     
  2. jcsd
  3. Dec 29, 2013 #2

    Mark44

    Staff: Mentor

    I don't see anything wrong in what you did.
    It's easy enough to check. Differentiate your solution implicitly and verify that it satisfies the differential equation.
     
  4. Dec 29, 2013 #3
    When you say [tex] ∫((1/x)y^2) dx = (y^2)ln|x| + C[/tex] you are assuming that ##y## is constant with respect to ##x##, contrary to the assumption of the problem that ##y## is a function of ##x##.

    If memory serves, the method of multiplying by an integrating factor in the manner that you have done works for first-order linear differential equations; ones that can be written in the form ##y'+f(x)y=g(x)##. The existence of a ##y^2## term means that this DE is not linear. So integrating factors may not be the way to go.
     
  5. Dec 29, 2013 #4
    gopher_p, you're correct, is this is nonlinear, so my method was incorrect. I needed to solve by substitution, see below:

    $$xy' = xy^{2} + y $$
    $$xy' - y = xy^{2} $$
    $$ \frac{xy'}{xy^2} - \frac{y}{xy^2} = 1 $$
    $$ \frac{y'}{y^2} - \frac{1}{xy} = 1 $$
    if $$ z = \frac{1}{y} $$.. and $$ z' = \frac{-1}{y^2} \frac{dy}{dx} $$
    then..
    $$ \frac{y'}{y^2} - \frac{1}{xy} = z' + \frac{1}{x}z $$
    this is now a linear equation..
    $$ z' + \frac{1}{x}z = -1 $$
    $$ e^{∫\frac{1}{x}} = x $$
    $$ ∫[x*z]' = ∫ -x dx $$
    $$ xz = -\frac{x^2}{2} + C $$
    $$ z = -\frac{x^2 + 2C}{2x} $$
    $$ \frac{1}{y} = -\frac{x^2+ 2C}{2x} $$
    $$ y = -\frac{2x}{x^2 + D} $$ where D = 2C
     
    Last edited: Dec 30, 2013
  6. Dec 29, 2013 #5
    Looks good. Nice work.
     
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