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:
- Is there any alternative I can use to solve the matrix ODE (equation (1))?
- How can I solve the third-order differential equation (equation (2))?
Thank you very much in advance!
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, $$ z(t)=z_0-c((y_0-C_3)t - C_4 + \sqrt{C_1}\sin(at + \frac{1}{2}b{t}^2 + C_2)) $$ $$ C_4=\sqrt{C_1}\sin(C_2) $$