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How could I solve this ode ?

$$({x^5}+{c^2}{x}) \frac{d^2 x}{d t^2}-\ 2{c^2}\Bigl(\frac{dx}{dt}\Bigr)^2-gc{x^3}=0$$

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1 Answer 1

up vote 4 down vote accepted

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.

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How could I solve v ???? the value of v is being ridiculous to solve the equation. –  ROBINSON Oct 17 '12 at 17:52
    
You'll get implicit solutions involving an integral that can't be done in closed form. If you're looking for an explicit closed-form formula for $x(t)$, I'm afraid you're not going to find one. –  Robert Israel Oct 17 '12 at 18:57

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