System of Linear Odes

$$ \begin{align*} \frac{dx}{dt}=a_{1}(t)x+b_{1}(t)y+f_{1}(t)\\ \frac{dy}{dt}=a_{2}(t)x+b_{2}(t)y+f_{2}(t) \end{align*} $$

Theorems #

General sol #

If homogeneous system has 2 linearly independent solutions $x_{1}(t),y_{1}(t)$ and $x_{2}(t),y_{2}(t)$ then general solution: $c_{1}x_{1}(t)+c_{2}x_{2}(t),c_{1}y_{1}(t)+c_{2}y_{2}(t)$

If non homogeneous solution has also a particular solution $x_{p}(t),y_{p}(t)$, then general solution: $c_{1}x_{1}(t)+c_{2}x_{2}(t)+x_{p}(t),c_{1}y_{1}(t)+c_{2}y_{2}(t)+y_{p}(t)$

derivative of wronskian #

$$ \begin{align*} \frac{d}{dt}W[t]&= \\ \\ &= [a_{1}+b_{2}]W \end{align*} $$

It is either 0 or “nowhere vanishing”.

Solving #

constant coeff #

We assume the solution of form $x(t)=Ae^{mt};y(t)=Be^{mt}$ and get an auxilary equation in terms of m

$$ \begin{align*} A(a_{1}-m)+B(b_{1})=0\\ A(a_{2})+B(b_{2}-m)=0 \end{align*} $$

$|Coeff-mI|=0 \Leftrightarrow m^{2}-(a_{1}+b_{2})m+(a_{1}b_{2}-a_{2}b_{1})=0$ for non-trivial solutions

different auxilary roots #

real #

If the auxiliary roots $m_{1},m_{2}$ are inequal, then we find 2 linear independent solutions $A_{1},B_{1}$ corresponding to $m_{1}$ $A_{2},B_{2}$ corresponding to $m_{2}$

imaginary #

If the auxilary root is imaginary, just consider one auxilary root and split the real and im parts to get 2 linearly independent solutions.

$$ \begin{align*} x=(A_{1}+iA_{2})e^{(a+ib)t}\\ y=(B_{1}+iB_{2})e^{(a+ib)t}\\ \\ x&= e^{at}((A_{1}\cos bt-A_{2}\sin bt)+i(A_{1}\sin bt+A_{2}\cos bt))\\ y&= a^{at}((B_{1}\cos bt-B_{2}\sin bt)+i(B_{1}\sin bt+B_{2}\cos bt)) \end{align*} $$

same auxiliary roots #

2nd solution would be of form

$\begin{cases}(A_{1}+A_{2}t)e^{mt} \(B_{1}+B_{2}t)e^{mt}\end{cases}$

$$ \begin{align*} (mA_{1}+A_{2}+mA_{2}t)e^{mt}=a_{1}(A_{1}+A_{2}t)e^{mt}+b_{1}(B_{1}+B_{2}t)e^{mt}\\ (mB_{1}+B_{2}+mB_{2}t)e^{mt}=a_{2}(A_{1}+A_{2}t)e^{mt}+b_{2}(B_{1}+B_{2}t)e^{mt}\\ \\ (a_{1}A_{2}+b_{1}B_{2}-mA_{2})t+(a_{1}A_{1}+b_{1}B_{1}-mA_{1}-A_{2})=0\\ (a_{2}A_{2}+b_{2}B_{2}-mB_{2})t+(a_{2}A_{1}+b_{2}B_{1}-mB_{1}-B_{2})=0\\ \\\ A_{2}(a_{1}-m)+B_{2}(b_{1})=0\\ A_{2}(a_{2})+B_{2}(b_{2}-m)=0\\ \\ A_{2},B_{2}=A,B \text{ (same as coeffs of the first sol)} \end{align*} $$

using matrix exponential #

$$ \begin{align*} \frac{dX}{dt}=AX\\ X: \text{col matrix }x(t),y(t)\\ A: \text{coeff matrix }\\ \\ S=e^{At} \end{align*} $$

S is a 2x2 matrix (C1 C2) where C1,C2 are the columns

It follows that:

$$ \begin{align*} \frac{dC_{1}}{dt}=AC_{1}\\ \frac{dC_{2}}{dt}=AC_{2}\\ \end{align*} $$ C1,C2 are 2 linearly independent solutions of X(t)