# Thread: Can't solve this Differential equation

1. ## Can't solve this Differential equation

Hey everyone, I've been trying to solve this differential equation for 2 days and I haven't gotten any closer to the answer. If someone could help that would be amazing, thank you. Here it is:
$\displaystyle xy' + y = y^2$
$\displaystyle y(1)=-11$

I've tried manipulating it to get y and dy on one side, and x and dx on the other side, and then integrating but that didn't work. I was left with:
$\displaystyle cx = y^2 - y$ where c is a constant. but i can't do anything from there.

I also tried doing $\displaystyle xy = \int y^2 dy$ where x is the constant of integration, but again, that didn't work out.

Thank you,
CenturionMonkey

2. $\displaystyle xy'+y={{y}^{2}}\implies \frac{y'}{{{y}^{2}}}+\frac{1}{xy}=\frac{1}{x}.$ Now put $\displaystyle u=\frac1y.$

3. $\displaystyle x^2(xy)'=x^2(xy'+y)=x^2y^2=(xy)^2.$ let $\displaystyle xy=u$ to get $\displaystyle \frac{u'}{u^2}=\frac{1}{x^2}.$ integrate both sides: $\displaystyle \frac{-1}{xy}=\frac{-1}{u}=\frac{-1}{x} + C.$ thus: $\displaystyle y=\frac{1}{1-Cx}.$ now use your initial condition to find $\displaystyle C.$

4. that actually kind of makes sense. Thanks a lot you two.

$\displaystyle x \frac{dy}{dx} = y^2 - y$
$\displaystyle \frac{dy}{y^2-y} = \frac{dx}{x}$.