given the first order system of differential equation

$$ \dot x (t) = -B(t)x(t)-C(t)y(t) $$

$$ \dot y(t) = A(t)x(t)+B(t)y(t) $$

can we always trasnform this first order system into a second order differential equation of the form

$$ \ddot x(t) = F( x(t), \dot x(t) , A(t) , B(t) , C(t) ) $$

or $$ \ddot y(t) = G( y(t), \dot x(t) , A(t) , B(t) , C(t) ) $$

i know that if there are a Hamiltonian system so $ H(x,y) $ is a Hamiltonian we can get the equations of motion by finding the Lagrangian but if there is no a Hamiltonian involved how could i do it


1 Answer 1


$\ddot x(t)=-B(t)\dot x(t)-\dot B(t)x(t)-C(t)\dot y(t)-\dot C(t)y(t)$

$\ddot x(t)=-B(t)\dot x(t)-\dot B(t)x(t)-C(t)(A(t)x(t)+B(t)y(t))-\dot C(t)y(t)$

$\ddot x(t)=-B(t)\dot x(t)-(\dot B(t)+A(t)C(t))x(t)-(\dot C(t)+B(t)C(t))y(t)$

$C(t)\ddot x(t)=-B(t)C(t)\dot x(t)-(\dot B(t)+A(t)C(t))C(t)x(t)-(\dot C(t)+B(t)C(t))C(t)y(t)$

$C(t)\ddot x(t)=-B(t)C(t)\dot x(t)-(\dot B(t)+A(t)C(t))C(t)x(t)+(\dot C(t)+B(t)C(t))(\dot x(t)+B(t)x(t))$

$C(t)\ddot x(t)=\dot C(t)\dot x(t)+(B(t)\dot C(t)-A(t)C(t)^2+B(t)^2C(t)-\dot B(t)C(t))x(t)$

$C(t)\ddot x(t)-\dot C(t)\dot x(t)-(B(t)\dot C(t)-A(t)C(t)^2+B(t)^2C(t)-\dot B(t)C(t))x(t)=0$


$\ddot y(t)=A(t)\dot x(t)+\dot A(t)x(t)+B(t)\dot y(t)+\dot B(t)y(t)$

$\ddot y(t)=A(t)(-B(t)x(t)-C(t)y(t))+\dot A(t)x(t)+B(t)\dot y(t)+\dot B(t)y(t)$

$\ddot y(t)=(\dot A(t)-A(t)B(t))x(t)+B(t)\dot y(t)+(\dot B(t)-A(t)C(t))y(t)$

$A(t)\ddot y(t)=(\dot A(t)-A(t)B(t))A(t)x(t)+A(t)B(t)\dot y(t)+(\dot B(t)-A(t)C(t))A(t)y(t)$

$A(t)\ddot y(t)=(\dot A(t)-A(t)B(t))(\dot y(t)-B(t)y(t))+A(t)B(t)\dot y(t)+(\dot B(t)-A(t)C(t))A(t)y(t)$

$A(t)\ddot y(t)=\dot A(t)\dot y(t)+(A(t)\dot B(t)+A(t)B(t)^2-\dot A(t)B(t)-A(t)^2C(t))y(t)$

$A(t)\ddot y(t)-\dot A(t)\dot y(t)-(A(t)\dot B(t)+A(t)B(t)^2-\dot A(t)B(t)-A(t)^2C(t))y(t)=0$


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