Is this the correct solution to this DiffEq?

  • Thread starter vanceEE
  • Start date
  • #1
109
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$$ xy' = y + xy^2 $$
$$ xy' - y = xy^2 $$
$$ y' + (-1/x)y = y^2 $$
$$ e^{∫(-1/x)dx} = 1/x $$
$$ (1/x)[y' + (-1/x)y = y^2] $$
$$ (1/x)y' - (1/x^2)y = (1/x)y^2 $$
$$ ∫((1/x)y)' = ∫((1/x)y^2) dx $$
$$ (1/x)y = (y^2)ln|x| + C $$
$$ y = (xy^2)ln|x|+Cx $$

Wolfram shows me another solution...
$$y = \frac{-2x}{C+x^2}$$
Are both correct?
 

Answers and Replies

  • #2
34,010
5,660
$$ xy' = y + xy^2 $$
$$ xy' - y = xy^2 $$
$$ y' + (-1/x)y = y^2 $$
$$ e^{∫(-1/x)dx} = 1/x $$
$$ (1/x)[y' + (-1/x)y = y^2] $$
$$ (1/x)y' - (1/x^2)y = (1/x)y^2 $$
$$ ∫((1/x)y)' = ∫((1/x)y^2) dx $$
$$ (1/x)y = (y^2)ln|x| + C $$
$$ y = (xy^2)ln|x|+Cx $$
I don't see anything wrong in what you did.
Wolfram shows me another solution...
$$y = \frac{-2x}{C+x^2}$$
Are both correct?
It's easy enough to check. Differentiate your solution implicitly and verify that it satisfies the differential equation.
 
  • #3
575
76
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.
 
  • #4
109
2
this DE is not linear.
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:
  • #5
575
76
Looks good. Nice work.
 

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