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I already asked a similar question on another post: Solving non linear differential equation.

\begin{align} &a \ \ddot{u}+b\ \left(\dot{u}\right)^2 +\dot{u}+\dot{u}\ c\ e^{u}+e^u-e^{2u}+1=0 \end{align}

How to solve this equation and what's the general solution? I tried to multiply by $\dot{u}$ but I was not able to solve the equation. Any suggestions?

Thanks.

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What is the independent variable of this ODE? –  doraemonpaul Jul 6 '12 at 21:39
    
$u=u(t)$, t is the independent variable –  Mark Jul 6 '12 at 22:10
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1 Answer

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$a~\ddot{u}+b\left(\dot{u}\right)^2+\dot{u}+\dot{u}ce^u+e^u-e^{2u}+1=0$

$a\dfrac{d^2u}{dt^2}+b\left(\dfrac{du}{dt}\right)^2+(ce^u+1)\dfrac{du}{dt}+e^u-e^{2u}+1=0$

Case $1$: $a\neq0$

This belongs to an ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0508.pdf

Let $v=\dfrac{du}{dt}$ ,

Then $\dfrac{d^2u}{dt^2}=\dfrac{dv}{dt}=\dfrac{dv}{du}\dfrac{du}{dt}=v\dfrac{dv}{du}$

$\therefore av\dfrac{dv}{du}+bv^2+(ce^u+1)v+e^u-e^{2u}+1=0$

$av\dfrac{dv}{du}=-bv^2-(ce^u+1)v+e^{2u}-e^u-1$

$v\dfrac{dv}{du}=-\dfrac{bv^2}{a}-\dfrac{(ce^u+1)v}{a}+\dfrac{e^{2u}-e^u-1}{a}$

This belongs to an Abel equation of the second kind.

In fact, all Abel equation of the second kind can be transformed into Abel equation of the first kind.

Let $v=\dfrac{1}{w}$,

Then $\dfrac{dv}{du}=-\dfrac{1}{w^2}\dfrac{dw}{du}$

$\therefore-\dfrac{1}{w^3}\dfrac{dw}{du}=-\dfrac{b}{aw^2}-\dfrac{ce^u+1}{aw}+\dfrac{e^{2u}-e^u-1}{a}$

$\dfrac{dw}{du}=-\dfrac{(e^{2u}-e^u-1)w^3}{a}+\dfrac{(ce^u+1)w^2}{a}+\dfrac{bw}{a}$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2

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