# Solve a homogeneous third-order differential equation with variable coefficients

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!

• 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? Mar 12, 2022 at 22:07
• 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 $b.\;$
– greg
Mar 14, 2022 at 3:23
• This problem is reminiscent of the Serret-Frenet equation but with time-dependent curvatures.
– greg
Mar 14, 2022 at 3:48
• 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. Mar 14, 2022 at 13:56

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, $$z(t)=z_0-c((y_0-C_3)t - C_4 + \sqrt{C_1}\sin(at + \frac{1}{2}b{t}^2 + C_2))$$ $$C_4=\sqrt{C_1}\sin(C_2)$$

• 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. Mar 13, 2022 at 3:16
• $+\tt1\,$ Choosing $C_1 = \sqrt{x_0^2+y_0^2}$ will make all subsequent equations less cluttered.
– greg
Mar 13, 2022 at 12:28
• It started wrongly unfortunately, i.e., $\frac{dy}{dt}$ is not $-(a+bt)x(t)$. Mar 13, 2022 at 15:23