I'm having trouble solving this ODE:
$$\ddot x = \mu \dot x^2 – g, \space \space x(0)=x_0$$
This is the ODE that determines the equation of motion of an object with air resistance. $\mu$ is a positive constant. This is what I've done:
Let $\dot x = v$ we have $v(0) = 0$ by assumption and the equation becomes:
$$\dot v = \mu v^2 – g$$
This is a Riccati differential equation. Recall that a Riccati differential equation has the form
$$\dot x = h(t) + f(t) x + g(t)x^2$$
In our case is $h(t) = -g$, $f(t) = 0$ and $g(t) = \mu$
To solve this kind of differential equations one has to guess a solution, which in this case I found be:
$$v_p(t) = \sqrt{g \over \mu}$$
$\implies y = {1 \over {v- \sqrt{g \over n}}}$ $\implies y(0) = \sqrt{n\over g}$ to find the solution $v(t)$ one has to solve the following differential equation:
$$\dot y = -(f(t) + 2v_pg(t))y-g(t)$$
This is what I've done:
\begin{align}
\dot y & = -(f(t) + 2v_pg(t))y-g(t) \\
& \dot y = -2 \sqrt{g\mu }\space y – \mu \space
& (1)
\end{align}
Now this is a linear inhomogeneous differential equation an the solution is given by the sum of the homogeneous solution $y_h(t)$ with the particular solution $y_p (t)$ The homogeneous solution comes from:
$$\dot y = -2 \sqrt{g\mu }\space y \implies y_h = y(0) e^{-2 \sqrt{g\mu }}$$
for the particular solution $y_p(t)$ is a little bit more complicated then to find it I substituted
$$y_p(t) = y(0)(t)e^{-2 \sqrt{g\mu }} = C_1 (t)e^{-2 \sqrt{g\mu }}$$
in the equation (1). This is what I get:
\begin{align}
y_p(t)& =C_1(t)' e^{-2 \sqrt{g\mu }} -2 \sqrt{g\mu }C_1(t)e^{-2 \sqrt{g\mu }} = -2 \sqrt{g\mu }C_1(t)e^{-2 \sqrt{g\mu }} – \mu
\end{align}
\begin{align}
\iff C_1(t)' e^{-2 \sqrt{g\mu }} & = -\mu
\end{align}
$$ \iff C_1(t) = {-\mu \over {2 \sqrt{g\mu }}}e^{2 \sqrt{g\mu }}$$
summarizing we have
$$y(t) = {1 \over v – \sqrt {g \over n}} = C_1(t)e^{-2 \sqrt{g\mu }} + y_h(t)= {-\mu \over {2 \sqrt{g\mu }}}e^{2 \sqrt{g\mu }}e^{-2 \sqrt{g\mu }} + y(0) e^{-2 \sqrt{g\mu }} $$
$$= {-\mu \over {2 \sqrt{g\mu }}} + y(0) e^{-2 \sqrt{g\mu }}$$
from here I tried to solve $y = {1 \over {v- \sqrt{g \over n}}}$ with respect to $v$ and then solve the last ODE $v = \dot x$ but I think that something has gone wrong.
Is there an alternative way to solve the original ODE? Is the method I've used the only one?
Best Answer
As you asked for methods I decided to post an alternative. Here we use the trick of eliminating the inhomogeneity.
First, consider only the equation for $v(t)=\dot{x}(t)$,
$$\dot{v}=\mu v^2-g.$$ The inhomogeneity $-g$ prevents us from simple integration of the ODE, so let's make it disappear by defining $w(t)=v(t)-\gamma$, which leads to
$$\dot{w}=\mu w^2+2\mu\gamma w+\mu\gamma^2-g.$$ We see that the choice $\gamma=\sqrt{g/\mu}$ kills the constant term, leadig to the homogenous equation for $w$: $$\frac{dw}{dt}=\mu w^2+2\sqrt{\mu g}w,$$ which now can directly be integrated:
$$\int_{w_0}^{w(t)} \frac{dw}{\mu w^2+2\sqrt{\mu g}w}=\int_0 ^t dt.$$ This can be solved using $\int dt/(at^2+bt)=-(2/b)\text{artanh} (2at/b+1)$ (for $b>0$) which gives
$$\frac{-1}{\sqrt{\mu g}}\text{artanh}\left( \mu w+1 \right)-C=t$$ with some integration constant $C$. We can solve this for $w$ and obtain $$w(t)=\sqrt{\frac{g}{\mu}}\left( \tanh[-\sqrt{\mu g}(t+C)]-1 \right).$$ Transforming back to our original function $v$ gives $$v(t)=w(t)+\gamma=\sqrt{\frac{g}{\mu}} \tanh[-\sqrt{\mu g}(t+C)],$$ and from the initial condition $v(0)=0$ it follows that $C=0$.
Now you can integrate $v=\dot{x}$ to get the position solution using that $\int dt \tanh(t)=\ln (\cosh x)$ arriving at
where the integration constant has been chosen such that $x(0)=x_0$.