Non-dimensional differential equation

[wel]

A body of constant mass is thrown vertically upwards from the ground. It can be shown that the appropriate non-dimensional differential equation for the height $x(t;u)$, reached at time $t\geq0$ is given by
\begin{equation} \frac{d^2x}{dt^2} = -1-\mu (\frac{dx}{dt})
\end{equation}
with corresponding initial conditions $x(0)=0, \frac{dx}{dt}(0) =1$, and where $0<\mu<<1.$
Deduce that the (non-dimensional) height at the highest point (where $\frac{dx}{dt} =0$) is given by
\begin{equation} h(\mu)= \frac{1}{\mu}- \frac{1}{\mu^2} log_e(1+\mu) \end{equation}

=>
It really hard for me to start

I was thinking do integration twice by doing the separation of variable:
\begin{equation} \frac{d^2x}{dt^2} = -1-\mu (\frac{dx}{dt})
\end{equation}
I got the general solution of \begin{equation}x(t)= \frac{log(t\mu +1) -t \mu}{\mu^2}\end{equation}

after that I do not know how to get the answer.
Please help me.

 

[LCKurtz]

A body of constant mass is thrown vertically upwards from the ground. It can be shown that the appropriate non-dimensional differential equation for the height $x(t;u)$, reached at time $t\geq0$ is given by
\begin{equation} \frac{d^2x}{dt^2} = -1-\mu (\frac{dx}{dt})
\end{equation}
with corresponding initial conditions $x(0)=0, \frac{dx}{dt}(0) =1$, and where $0<\mu<<1.$
Deduce that the (non-dimensional) height at the highest point (where $\frac{dx}{dt} =0$) is given by
\begin{equation} h(\mu)= \frac{1}{\mu}- \frac{1}{\mu^2} log_e(1+\mu) \end{equation}

=>
It really hard for me to start

I was thinking do integration twice by doing the separation of variable:
\begin{equation} \frac{d^2x}{dt^2} = -1-\mu (\frac{dx}{dt})
\end{equation}
I got the general solution of \begin{equation}x(t)= \frac{log(t\mu +1) -t \mu}{\mu^2}\end{equation}

after that I do not know how to get the answer.
Please help me.

How did you get that solution? You have a constant coefficient second order DE. You wouldn't expect a logarithm in the solution.