Non-dimensional differential equation 1

[ra_forever8]

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.

 

[I like Serena]

Re: Non-dimensional differential equation

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.

Your method is the right approach.

Let me show you the first step to find $v$.

Let $v=\frac {dx}{dt}$ and let $\dot v = \frac {d^2x}{dt^2}$.
Then:
\begin{array}{}
\dot v &=& -1 - \mu v \\
\frac{\dot v}{1 + \mu v} &=& -1 \\
\frac 1 \mu \ln(1+\mu v) &=& -t + C \\
v &=& C' e^{-\mu t} - \frac 1 \mu \\
\end{array}

Applying the boundary condition $v(0)=1$ yields:
$$v = \frac 1 \mu((\mu + 1)e^{-\mu t} - 1)$$

From here you can find the time $t$ at which $v=0$.
And you can also integrate again to find $x$.