Originally Posted by

**chisigma** The second order DE has the form...

$\displaystyle x^{''} = f(x)$ (1)

... where...

$\displaystyle f(x)= \frac{k}{x-c}$ (2)

The approach to solve (1) is the following: because is...

$\displaystyle x^{''} = \frac{d x^{'}}{dt} = \frac{d x^{'}}{dx}\cdot \frac{dx}{dt} = x^{'}\cdot \frac{dx^{'}}{dx}$ (3)

... the (1) becomes...

$\displaystyle x^{'}\cdot \frac{dx^{'}}{dx}= f(x) \rightarrow x^{'}\cdot dx^{'} = f(x)\cdot dx$ (4)

... so that we have a first order DE in $\displaystyle x^{'}$ and $\displaystyle x$. Integrating (4) we obtain...

$\displaystyle \frac{x^{'2}}{2} = \phi (x) + c_{1} \rightarrow x^{'} = \sqrt{2\cdot \phi(x) + c_{1}} $ (5)

... where $\displaystyle \phi(x)$ is a primitive of $\displaystyle f(x)$. Separating variables again in (5) we obtain...

$\displaystyle dt = \frac{dx}{ \sqrt{2\cdot \phi(x) + c_{1}}} $ (6)

... so that is...

$\displaystyle t= \int \frac{dx}{ \sqrt{2\cdot \phi(x) + c_{1}}} + c_{2}$ (7)

Now what remains to do is compute a primitive of $\displaystyle f(x)$ given by (2), insert it in (7), and then integrating again... very easy (Wink) ...

Kind regards

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