Variable Separable Form for $\frac{dR}{dt}=\frac{t^3+r^3}{t^2r+tr^2}$

Question

I just need a hint or a method on how to separate the equation $$\frac{dR}{dt}=\frac{t^3+R^3}{t^2R+tR^2}$$ so I can integrate it and get a general solution in terms of $R(t)$

Answer 1

$$\frac{dR}{dt}=\frac{t^3+R^3}{t^2R+tR^2}$$ $$\frac{dR}{dt}=\frac{(t^2-tR+R^2)}{tR}$$ $$\frac{dR}{dt}=\dfrac tR-1+\dfrac Rt$$ It's homogeneous. $R=ut \implies R'=u't+u$ $$u't+u=\dfrac 1u-1+u$$ $$u't=\dfrac 1u-1$$ It's separable.

Answer 2

Let $v=\frac{R}{t}\Rightarrow R=vt$ . Then, differentiating both sides of this equation w.r.t. t, we get, $$\frac{dR}{dt}=v+t\frac{dv}{dt}$$ Now, dividing the numerator and denominator of the right hand side of the given differential equation, we get,

$$\frac{dR}{dt}=\frac{1+\displaystyle{\frac{R^3}{t^3}}}{\displaystyle{\frac{R}{t}+\frac{R^2}{t^2}}}=\frac{1+v^3}{v+v^2}\Rightarrow v+t\frac{dv}{dt}=\frac{1+v^3}{v+v^2}$$
$$\Rightarrow t\frac{dv}{dt}=\frac{1+v^3}{v+v^2}-v=\frac{1-v^2}{v+v^2}$$ $$\Rightarrow \frac{v+v^2}{1-v^2}dv=\frac{dt}{t}$$ And so the equation becomes separable in $v$ and $x$.