Consider an ODE of the form $$\frac{d^2x}{dt^2}+p(t)\frac{dx}{dt}+q(t)x=0$$ Suppose that we have two solutions $x_1(t)$ and $x_2(t)$ to this ODE and their Wronskian is a non-zero constant. That is, $$W[x_1,x_2]=C\neq0$$ What is $p(t)$? (Find an explicit formula).
The formula for the Wronskian is $$W[x_1,x_2]=x_1\frac{dx_2}{dt}-x_2\frac{dx_1}{dt}$$ Since the Wronskian is non-zero, we know that $x_1$ and $x_2$ are independent. This means that $\frac{dx_1}{dt}$ and $\frac{dx_2}{dt}$ are such that all variables cancel out. I'm not sure how to proceed from here. Can I get some hints on how to find $p(t)$? Thanks
Edit: Using Abel's theorem, we know that if $x_1$ and $x_2$ are solutions to the ODE then $$W[x_1,x_2]=Ae^{-\int p(t)dt}$$ Then this can be used to solve for $p(t)$. \begin{align} W[x_1,x_2]&=Ae^{-\int p(t)dt}=C\\ &e^{-\int p(t)dt}=\frac{C}{A}\\ &\int p(t)dt=-\ln(\frac{C}{A})\\ &p(t)=\frac{d}{dt}\left(-\ln(\frac{C}{A})\right)\\ & =0 \end{align} So $p(t)=0$ Can I get some verification on this? Thank you
If $x_i''+px_i'+qx_i=0$, $i=1,2$, then $$ x_1(x_2''+px_2'+qx_2)=x_2(x_1''+px_1'+qx_1)=0 \\ \Longrightarrow\quad (x_1x_2''-x_1''x_2)+p(x_1x_2'-x_2x_1')=0 \\ \Longrightarrow\quad p=-\frac{x_1x_2''-x_1''x_2}{x_1x_2'-x_2x_1'}=-\frac{x_1x_2''-x_1''x_2}{W[x_1,x_2]} $$