(Context) : In physics class, we learned how to derive a differential equation of the movement of a system (given the context of what is studied) I tried applying these methods to derive the differential equations of the movement of a paper airplane (weight, lift, drag and nose-up force)
But I got 2 differential equations that are intertwined :
$X''(t) = -\frac{K_1}{m}X'(t)-\frac{K_3}{m}Y'(t)$
$Y''(t) = -g-\frac{K_1}{m}Y'(t)+\frac{K_2+K_3}{m}X'(t)$
Where $K_1$, $K_2$, $K_3$, $g$ and $m$ are constants.
I really don't know how to solve these type of differential equations, and couldn't find anything online that could help me much.
If anyone could give the method to get $X(t)$ and $Y(t)$, I would be glad to hear it.
BTW if needed in any ways, the initial conditions are :
- $X(0)=x_0$
- $Y(0)=y_0$
- $X'(0)=\cos(\theta)V_0$
- $Y'(0)=\sin(\theta)V_0$
The idea is to differentiate and substitute to obtain a differential equation in just one variable. Rearranging the first equation, \begin{align*} Y'(t) &= -\frac{K_1}{K_3} X'(t) - \frac{m}{K_3} X''(t) \\ \implies Y''(t) &= -\frac{K_1}{K_3} X''(t) - \frac{m}{K_3}X'''(t). \end{align*} Substituting these into the second equation, \begin{align*} &-\frac{K_1}{K_3} X''(t) - \frac{m}{K_3} X'''(t) = -g - \frac{K_1}{m}\left(-\frac{K_1}{K_3} X'(t) - \frac{m}{K_3} X''(t) \right) + \frac{K_2 + K_3}{m}X'(t) \\ \implies&\frac{m}{K_3} X'''(t) + \frac{2K_1}{K_3} X''(t) + \left(\frac{K_1^2}{mK_3} + \frac{K_2+ K_3}{m}\right)X'(t) -g =0 \\ \implies&\frac{m}{K_3} x''(t) + \frac{2K_1}{K_3} x'(t) + \left(\frac{K_1^2}{mK_3} + \frac{K_2+ K_3}{m}\right)x(t) -g =0, \end{align*} where $x(t) = X'(t)$. This is now a standard homogeneous second-order linear ODE in $x$, and (technically) easy to solve, although I can't imagine the solution is pretty. Finding $X$ is then easy by integrating once.
To find $Y$, either use the same process to form an ODE in $Y$, or substitute $X'$ and $X''$ into the first equation.