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I'm stuck on the following problem: given the two ODE: $$\ddot{x}(t)=-\beta \sqrt{\dot{x}(t)^2+\dot{y}(t)^2}\dot{x}(t)$$ $$\ddot{y}(t)=-g+\beta \sqrt{\dot{x}(t)^2+\dot{y}(t)^2}\dot{y}(t)$$ in which $g$ and $\beta$ are constant, what is the solution of the system of ODE, knowing the initial conditions: $$x(0)=x_0,y(0)=y_0$$ $$\dot{x}(0)=x_1,\dot{y}(0)=y_1$$ at time $t=0$? Thanks.

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If you multiply the first equation by $\dot{y}$ and the second by $\dot{x}$, you can find a first integral by adding them, leaving $$ \dot{x}(t)\dot{y}(t) = -g\big(x(t)-x_0\big) + x_1 y_1$$ Maybe that helps. –  Pragabhava Nov 1 '12 at 3:04
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up vote 1 down vote accepted

I do not think you can get a closed form solution for your system. However, you can get a series solution

$$ x(t) = x_{{0}}+x_{{1}}t - \frac{1}{2}\,\beta\,\sqrt {{x_{ {1}}}^{2}+{y_{{1}}}^{2}}\,x_{{1}}{t}^{2}+ O ( {t}^{3} ),$$

$$ y( t ) = y_{{0}}+y_{{1}}t+ \left( -\frac{1}{2}\,g+\frac{1}{2}\,\beta\, \sqrt {{x_{{1}}}^{2}+{y_{{1}}}^{2}}y_{{1}} \right) {t}^{2}+O ( t ^{3} )\,.$$

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Suppose $\beta\neq0$ :

Let $\begin{cases}u(t)=\dot{x}(t)\\v(t)=\dot{y}(t)\end{cases}$ ,

Then $\begin{cases}\dot{u}(t)=\ddot{x}(t)\\\dot{v}(t)=\ddot{y}(t)\end{cases}$

$\therefore\begin{cases}\dot{u}(t)=-\beta\sqrt{u(t)^2+v(t)^2}u(t)~......(1)\\\dot{v}(t)=-g+\beta\sqrt{u(t)^2+v(t)^2}v(t)~......(2)\end{cases}$

$\dfrac{(2)}{(1)}$ : $\dfrac{dv}{du}=\dfrac{-g+\beta\sqrt{u^2+v^2}v}{-\beta\sqrt{u^2+v^2}u}=\dfrac{g}{\beta\sqrt{u^2+v^2}u}-\dfrac{v}{u}$

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