Help finding an explicit solution to this ODE

Question

I have the following ODE:

$$\dot{r}=(r-r^{3})\sin^{2}(\theta)$$

I have been trying to find an explicit solution for $r$, with little success. I have tried separating variables but then get stuck trying to make it explicit. Any suggestions? Thanks.

Answer 1

$$\frac{dr}{d\theta}=(r-r^3)\sin^2\theta$$ Separating the variables: $$\int\frac{1}{r-r^3}dr=\int\sin^2\theta d\theta$$ First, let's focus on the integral on the left hand side. $$\begin{align} \int\frac{1}{r-r^3}dr&=\int\frac{1}{r(1-r)(1+r)}dr=\int\frac{1}{r}+\frac{1}{2(1-r)}-\frac{1}{2(1+r)}dr\\ &=\ln\lvert r\rvert-\frac{1}{2}\ln\lvert1-r\rvert-\frac{1}{2}\ln\lvert1+r\rvert+C\\ &=\frac{1}{2}\ln\left\lvert\frac{r^2}{1-r^2}\right\rvert+C\\ &=\frac{1}{2}\ln\left\lvert\frac{r^2-1+1}{1-r^2}\right\rvert+C=\ln\left\lvert\frac{1}{1-r^2}-1\right\rvert+C\end{align}$$ Now for the right integral. $$\int\sin^2\theta d\theta=\frac{1}{2}\int1+\cos 2\theta d\theta=\frac{1}{2}\theta+\frac{1}{4}\sin2\theta+D$$ Hence, $$\ln\left\lvert\frac{1}{1-r^2}-1\right\rvert=\theta+\frac{1}{2}\sin2\theta+C_1$$ And now we can rearrange to find $r^2$, and thus find $r$.

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I hope that helps. If you have any questions please don't hesitate to ask :)