Solving a system of eqns. by variation of parameters - When the Wronskian is 0
$$y''-2y'+y=3e^{-t}$$ Rewrite it as: $$(y(t)e^{-t})''=3e^{-2t}$$ Integrate twice.
Note that the Wronskian is not zero: $$y_1(t)=e^t \implies y_2(t)=te^t$$ $$W(y_1,y_2)=W(e^t,te^t)$$ $$W(e^t,te^t)=e^t(e^t+te^t)-te^{2y}=e^{2t}$$