Consider the linear ODE $y''+py'+qy=0$, where p and q are constants
1). Convert this second-order into a planar system
I got $v=y'$ and $v'=-pv-qy$
2). Write the planar system in matrix form, so that y'=Ay and find the characteristic polynomial of A.
Question1 : I am not quite sure how to get the matrix.
3).Solving the ODE y''-4y=0 by converting this ODE into a planar system, then construct the general solution for the planar system. Finally, extract a general solution to the original ODE from your solution to the planar system.
Question 2: I got $v'=4y$ and $y'=v$, what am I supposed to do then?
Best Answer
For the first question : $$y''+py'+qy=0$$ $$u=y , v=y'$$ $$\pmatrix {u \\ v}'=\pmatrix {0 & 1 \\ -q & -p} \pmatrix {u \\ v}$$ For the differential equation : $$ y''-4y=0 $$ With matrices notation: $$\pmatrix {u \\ v}'=\pmatrix {0 & 1 \\ 4& 0} \pmatrix {u \\ v}$$ For the polynomial characteristic: $$p(\lambda)=\begin{vmatrix} -\lambda & 1\\ 4 & -\lambda \end{vmatrix} $$ $$p(\lambda)=\lambda ^2-4$$