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I have a system of first order differential equations:

$\begin{bmatrix}x'(t) \\ y'(t) \\ z'(t)\end{bmatrix} = \begin{bmatrix} 0 & a+bt & 0 \\ -(a+bt) & 0 & c \\ 0 & -c & 0\end{bmatrix} \begin{bmatrix}x(t) \\ y(t) \\ z(t)\end{bmatrix}, \;\;\;\;\;\;\;\;\;\;\;\;(1)$

where $a$, $b$, and $c$ are constants. The above matrix ODE can be reduced to a homogeneous third-order differential equation with non-constant coefficients:

$y'''(t) + \left((a+bt)^2+c^2\right)y'(t) +3b(a+bt)y(t) = 0. \;\;\;\;\;\;\;\;\;\;\;\;(2)$

Solving $y(t)$ from the above equation can further solve $x(t)$ and $z(t)$. However, I am kind of stuck at this point. So my questions are:

  1. Is there any alternative I can use to solve the matrix ODE (equation (1))?
  2. How can I solve the third-order differential equation (equation (2))?

Thank you very much in advance!

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  • $\begingroup$ I found a mistake previously and just update the question. I would say that the primary question would be, whether it is possible to have an analytical expression for the solution. If so, what is it? $\endgroup$
    – chchien
    Mar 12, 2022 at 22:07
  • $\begingroup$ Multiplying the matrix ODE by $x^T$ yields $$\eqalign{ x^T\dot x &= x^T(Mx) \;=\; 0 \qquad \{{\rm since}\,M\,{\rm is\,skew}\} \\ }$$ Therefore the position vector is always orthogonal to its velocity. This implies some sort of circular motion with angular velocity $b.\;$ $\endgroup$
    – greg
    Mar 14, 2022 at 3:23
  • $\begingroup$ This problem is reminiscent of the Serret-Frenet equation but with time-dependent curvatures. $\endgroup$
    – greg
    Mar 14, 2022 at 3:48
  • $\begingroup$ Yes, I was trying to solve the Frenet-frame with an assumption that the curvature is modeled by a first order linear equation while the torsion is assumed to be constant. If curvature and torsion are constant, the problem is easy, but for this case, I wonder whether an analytic solution exists. $\endgroup$
    – chchien
    Mar 14, 2022 at 13:56

1 Answer 1

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We have, $$ \frac{dx}{dt}=(a +bt)y(t) $$ $$ \frac{dy}{dt}=-(a +bt)x(t) $$ $$ \frac{dz}{dt}=-cy(t) $$ Take the ratio of the first two equations, $$ \frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}=-\frac{x}{y} $$ Integrate to get $y$ in terms of $x$, $$ \int_{y0}^y y^{'}dy^{'}=-\int_{x_0}^x x^{'}dx^{'} $$ $$ y^2=-x^2 +y_0^2+x_0^2 $$ with $C_1=y_0^2+x_0^2$ and taking the positive branch, $$ y=\sqrt{C_1-x^2} $$ $$ dx=(a+bt)\sqrt{C_1-x^2}dt $$ $$ \int_{x_0}^x \frac{dx^{'}}{\sqrt{C_1-{x^{'}}^2}}=\int_0^t (a+bt^{'})dt^{'} $$ resulting in, $$ \sin^{-1}(\frac{x}{\sqrt{C_1}})=at + \frac{1}{2}bt^2 +\sin^{-1}(\frac{x_0}{\sqrt{C_1}}) $$ $$ x(t)=\sqrt{C_1}\sin(at + \frac{1}{2}bt^2 + C_2) $$ $$ C_2=\sin^{-1}(\frac{x_0}{\sqrt{C_1}}) $$ We now have from the second d.e. $$ \int_{y0}^y dy^{'}=-\int_0^t (a+bt^{'})\sqrt{C_1}\sin(at^{'} + \frac{1}{2}b{t^{'}}^2 + C_2)dt^{'} $$ We make the substitution $u^2=at + \frac{1}{2}b{t}^2 + C_2$ with $dt=\frac{2u}{a+bt}$ with the result $$ y=y_0 -C_3 +\sqrt{C_1}\cos(at^{'} + \frac{1}{2}b{t^{'}}^2 + C_2) $$ $$ C_3=\sqrt{C_1}\cos( C_2) $$ For the $z$ derivative we have $$ \int_{z_0}^z dz^{'}=-c\int_0^t (y_0 -C_3 +\sqrt{C_1}\cos(at^{'} + \frac{1}{2}b{t^{'}}^2 + C_2))dt^{'} $$ Using the same substitution as before we find, $$ z(t)=z_0-c((y_0-C_3)t - C_4 + \sqrt{C_1}\sin(at + \frac{1}{2}b{t}^2 + C_2)) $$ $$ C_4=\sqrt{C_1}\sin(C_2) $$

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  • $\begingroup$ Thank you so much! However, I have three questions: (1) $\frac{dy}{dt}=-(a+bt)x(t)+cz(t)$, not, $\frac{dy}{dt}=-(a+bt)x(t)$. (2) Should the substitution be $u=at+\frac{1}{2}bt^2+C_2$ and $dt=\frac{du}{a+bt}$? (3) The last part that solves z(t), the same substitution cannot be applied because there is no $a+bt$ inside the integral. I hope I am not making a mistake. $\endgroup$
    – chchien
    Mar 13, 2022 at 3:16
  • $\begingroup$ $+\tt1\,$ Choosing $C_1 = \sqrt{x_0^2+y_0^2}$ will make all subsequent equations less cluttered. $\endgroup$
    – greg
    Mar 13, 2022 at 12:28
  • $\begingroup$ It started wrongly unfortunately, i.e., $\frac{dy}{dt}$ is not $-(a+bt)x(t)$. $\endgroup$
    – chchien
    Mar 13, 2022 at 15:23

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