Recent content by blizzard750

y1(t) = t^(2)+5t, y2(t) = t^(2)-5t I know that these functions are linearly independent because they are not scalar multiples, but every time that i do the Wronskian i get 0. [t^(2) + 5t t^(2)-5t] [2t+5 2t -5] (2t^(3) -25t) - (2t^3-25t) = 0 my instructor some how...

y''-2y'-3y=-3te^-t i know that that the general solution to this problem is yh = c1e^3t + c2e^-t i am having trouble figuring out what the particular solution is (yp) i keep getting the yp = 3/4te^-t , but wolfram alpha is telling me that the answer is something else. how do i get...