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**chisigma** The DE is of the type $\displaystyle x^{''}= f(x)$ and can be reduced to a first order DE as...

$\displaystyle \displaystyle x^{'}\ d x^{'}= f(x)\ dx = \{\frac{k_{2}}{x} - \frac{k_{1}}{(x-a)^{2}} \}\ dx$ (1)

... where the variables $\displaystyle x^{'}$ and $\displaystyle x$ are separated. Integrating (1) in standard fashion You obtain...

$\displaystyle \displaystyle x^{'}= \pm \sqrt{\frac{2 k_{1}}{x-a} - \frac{2 k_{2}}{x} + c_{1}} $ (2)

Now the 'initial conditions' say that is $\displaystyle x^{'} (\frac{a}{2})=0$ so that is...

$\displaystyle \displaystyle c_{1}= \frac{4 (k_{1}+ k_{2})}{a}$ (3)

... and $\displaystyle x^{'} $ as function of $\displaystyle x$ is found. Now are You able to proceed?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$