$$ x\frac{dy}{dx} = y(y+1) $$
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?
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$\begingroup$ You'd get $\ln x+\ln c= \ln y-\ln{(y+1)}\implies \frac {y}{y+1}=cx$. Now solve for $y$. Thus $y(1-cx)=cx$. $\endgroup$– PNDasCommented Dec 15, 2020 at 18:38
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$\begingroup$ I got lost in the step where you said solve for y. But I get what you said before. $\endgroup$– draw electricCommented Dec 15, 2020 at 18:51
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$\begingroup$ In the future, try and typeset your mathematics using MathJax. $\endgroup$– C.M. LeesCommented Dec 15, 2020 at 19:25
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2 Answers
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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.
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$\begingroup$ I found out the question is really weird. To solve it use y = xe^c(y+1) (Transfer In(y+1) to the other side) . Expand, subtract the value xy^c. Now you can factorize out they to get the answer. The principle relies on multiplying the other side by y so you can subtract and factorize. $\endgroup$ Commented Dec 15, 2020 at 20:12
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\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}
answered Dec 15, 2020 at 18:44