Differential equation - solving for y.

$$x\frac{dy}{dx} = y(y+1)$$

I'm unsure how to get rid of the two $$y$$'s I get further down.

I did $$1/x\ dx = 1/ ( y^2+y )$$. After partial fractions $$\ln x = \ln y - \ln (y - 1)$$

I have two $$y$$'s and I don't know how to get $$y$$ by itself. Help?

• You'd get $\ln x+\ln c= \ln y-\ln{(y+1)}\implies \frac {y}{y+1}=cx$. Now solve for $y$. Thus $y(1-cx)=cx$. Commented Dec 15, 2020 at 18:38
• I got lost in the step where you said solve for y. But I get what you said before. Commented Dec 15, 2020 at 18:51
• In the future, try and typeset your mathematics using MathJax. Commented Dec 15, 2020 at 19:25

Here's how I'd go about solving this ODE. \begin{align} x\frac{dy}{dx} &= y(y+1)\tag*{[given]} \\ \int\frac{dy}{y(y+1)} &= \int\frac{dx}{x}\tag*{[separate variables & integrate]} \\ \int\left(\frac{1}{y}-\frac{1}{y+1}\right)\,dy &= \int\frac{1}{x}\,dx\tag*{[partial fractions on LHS]} \\ \ln{|y|}-\ln{|y+1|} &= \ln{|x|}+C_1,\ C_1\in\mathbb{R}\tag*{[perform integration]} \\ \ln{\left|\frac{y}{y+1}\right|} &= \ln{|x|}+C\tag*{[simplify LHS]} \\ \left|\frac{y}{y+1}\right| &= \exp\left(\ln{|x|}+C\right)\tag*{[exponentiate]} \\ &= C_2|x|,\ C_2\in\mathbb{R}\tag*{[simplify RHS]} \\ &= C_2x,\ x\geq 0.\tag*{[choose x\geq 0]} \end{align} It doesn't look like there's a convenient way to express $$y$$ as an explicit function of $$x$$, so that's as far as I'll go. I hope this helps.
\begin{align} x \frac{dy}{dx} &= y(y+1) \\ \frac{dy}{y(y+1)} &= \frac{dx}{x} \\ \frac{dy}{y} - \frac{dy}{y+1} &= \frac{dx}{x} \\ &\phantom{n}\vdots \end{align}