# Is this the correct solution to this DiffEq?

1. Dec 29, 2013

### vanceEE

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

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

2. Dec 29, 2013

### 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.

3. Dec 29, 2013

### gopher_p

When you say $$∫((1/x)y^2) dx = (y^2)ln|x| + C$$ 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. Dec 29, 2013

### vanceEE

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
5. Dec 29, 2013

### gopher_p

Looks good. Nice work.