System of Differential Equations: Accelerating Point

Question

At $t=0$, a point starts at $(10,10)$ with a velocity of $20$ units per second in the positive $x$ direction. It also has a constant acceleration of $5$ units per second squared towards to origin at all times. What is its parametric equation?

First let me say that I am very unexperienced when it comes to differential equations, and I have no idea how to even approach this one.

To solve this problem, I started by saying that, if both $x$ and $y$ are functions of $t$, $$x''=5\cos\arctan\frac{y}{x}=\frac{5x}{\sqrt{x^2+y^2}}$$ $$y''=5\sin\arctan\frac{y}{x}=\frac{5y}{\sqrt{x^2+y^2}}$$

Furthermore, because of the starting velocity, when $x''$ is integrated, the constant will end up being $20$. And when $x'$ and $y'$ are integrated, due to the starting position of the point, their constants will both be $10$. However, I don't know how to get to that point... I only know how to integrate single differential equations, and only simple ones at that. Can anybody explain to me if and how one can go about solving this problem?

Be warned: I made up this problem and I have no idea if it has a neat solution. If it can't be done and I am wasting my time, just tell me and explain why.

Answer 1

In polar coordinates, the equations of movement are:

$$\begin{cases} \ddot r-r\dot\theta=-5\\ \dfrac{1}{r}\dfrac{d(r^2\dot\theta)}{dt}=0\implies r^2\dot\theta=L_\theta \end{cases}$$

$$\ddot r-L_\theta/r=-5$$

It has not a simple solution... Elliptical integrals are needed.

what would a planetary orbit look like if gravity had constant magnitude?

Answer 2

The Lagrangian of a particle subjected to the constant acceleration towards the origin written in plane polar is $$\mathcal{L}(r, \varphi, \dot{\varphi}, \dot{r})=\frac{1}{2}(\dot{r}^{2}+r^{2}\dot{\varphi}^{2})+\alpha{r}$$ Where $\alpha$ is the magnitude of the acceleration. $\varphi$ is a cyclic variable, the conserved angular momentum is $$L_{\varphi}=\frac{\partial\mathcal{L}}{\partial\dot{\varphi}}=r^{2}\dot{\varphi}$$ So, $\dot{\varphi}=\frac{L_{\varphi}}{r^{2}}$, and the lagrnagian becomes $$\mathcal{L}(r, \dot{r})=\frac{1}{2}\dot{r}^{2}+\frac{L_{\varphi}^{2}}{r^{2}}+\alpha{r}$$ The First integral is
$$E=\dot{r}\frac{\partial{\mathcal{L}}}{\partial{\dot{r}}}-\mathcal{L}=$$ $$=\frac{1}{2}\dot{r}^{2}-\frac{L_{\varphi}^{2}}{r^{2}}-\alpha{r}$$ Thus $$\dot{r}=\pm\sqrt{2E+\frac{2L_{\varphi}^{2}}{r^{2}}+2\alpha{r}}$$ or $$t+c=\pm\int\frac{dr}{\sqrt{2E+\frac{2L_{\varphi}^{2}}{r^{2}}+2\alpha{r}}}$$ From this integral (if you manage to do it, via some elliptic functions e.g.), you can find $r(t)$ by inversion, the other parametric equation, for $\varphi$ is obtained by $$\varphi+\varphi_{0}=\int\frac{L_{\varphi}}{r^{2}(t)}dt$$ The constants $\varphi_{0}$, $L_{\varphi}$, $E$ and $t_{0}$ are all found from the initial datum E.g. in your sepecific problem $\varphi_{0}=\frac{\pi}{4}$ and $t_{0}=0$. As the angular momentum is an integral of motion it is equal to the value it has initially at all times. $L_{\varphi}=r_{0}^{2}\dot{\varphi}_{0}$. $r_{0}^{2}=x_{0}^{2}+y_{0}^{2}=200$. The initial velocity in the $x$ diraction is $$\dot{x}=\dot{r}_{0}\cos{\varphi_{0}}-r_{0}\sin{\varphi_{0}}\dot{\varphi_{0}}=20$$ and $$\dot{y}_{0}=\dot{r}_{0}\sin{\varphi_{0}}+r_{0}\cos{\varphi_{0}}\dot{\varphi_{0}}=0$$ So, $$\dot{\varphi}_{0}=-\frac{1}{10}$$ And finally $L_{\varphi}=-20$