How to solve this first order non-linear ODE

Question

I am struggling to solve this first order ODE.

$$ u'\,^2 = a u^2 + b u^3$$

Mathematica gives me,

$$ u(v) = \dfrac{a}{b} \mathrm{sech} \left( \dfrac{1}{2} \sqrt{a} \ (v + c) \right)^2 $$

So I assume there is some nice method I can use to get this, any ideas?

Answer 1

Hint

You could notice that the equation is separable and that you can write it as $$\frac{dv}{du}=\frac{1}{\sqrt{a u^2+b u^3}}$$ and the integration leads to $$v=-\frac{2 \tanh ^{-1}\left(\sqrt{\frac{a+b u}{a}}\right)}{\sqrt{a}}+c$$ I am sure that you can take from here.

Added later to this answer

All of that means that if the following change of variable is made $$u=\frac{a \left(z^2-1\right)}{b}$$ after simplification, we have $$\frac{dv}{dz}=-\frac{2}{\sqrt{a} \left(z^2-1\right)}$$ and then $$v=\frac{2 \tanh ^{-1}(z)}{\sqrt{a}}+c$$

Answer 2

Rewrite: $$ \begin{align} u'^2&=au^2+bu^3\\ \left(\frac{du}{dv}\right)^2&=u^2(a+bu)\\ \frac{du}{dv}&=u\sqrt{a+bu}\\ dv&=\frac{du}{u\sqrt{a+bu}}\\ \int\ dv&=\int\frac{1}{u\sqrt{a+bu}}\ du\\ &=\int\frac1a\left[\frac{\sqrt{a+bu}}{u}-\frac{b}{\sqrt{a+bu}}\right]\ du. \end{align} $$ For the first integral in the RHS, let $x^2=a+bu$ and for the second integral in the RHS, let $y=a+bu$, then \begin{align} \int\ dv&=\frac{2}{a}\int\frac{x^2}{x^2-a}\ dx-\frac{1}{a}\int y^{\large-\frac12}\ dy\\ &=\frac{1}{a}\int\left[\frac{x}{x-\sqrt{a}}+\frac{x}{x+\sqrt{a}}\right]\ dx-\frac{1}{a}\int y^{\large-\frac12}\ dy\\ \end{align} The last step should be easy and I will leave it to you.