How could I solve this ode ?
$({x^5}+{c^2}{x}) \frac{d^2 x}{d t^2}-\ 2{c^2}\Bigl(\frac{dx}{dt}\Bigr)^2-gc{x^3}=0$
How could I solve this ode ?
$({x^5}+{c^2}{x}) \frac{d^2 x}{d t^2}-\ 2{c^2}\Bigl(\frac{dx}{dt}\Bigr)^2-gc{x^3}=0$
This is an autonomous second-order differential equation, so we can write it as a first-order equation in $x$ and $v = dx/dt$: $ {\frac {dv}{dx}} ={\frac {2\,{c}^{2} v^{2}+gc{x}^{3}}{ \left( {x}^{5}+{c}^{2}x \right) v }} $ That is a Bernoulli differential equation. The change of variables $v = \pm\sqrt{u}$ transforms it to a linear differential equation ${\frac {d}{dx}}u \left( x \right) =4\,{\frac {{c}^{2}u \left( x \right) }{x \left( {x}^{4}+{c}^{2} \right) }}+2\,{\frac {gc{x}^{2}}{{ x}^{4}+{c}^{2}}} $ whose general solution is $ u \left( x \right) =-{\frac {2gc{x}^{3}}{{x}^{4}+{c}^{2}}}+A {\frac {{ x}^{4}}{{x}^{4}+{c}^{2}}} $ where $A$ is an arbitrary constant.