Hassan H. answered • 07/29/13

Math Tutor (All Levels)

Hello Sun,

The problem you pose is probably best handled by the method of Variation of Parameters. We first seek to solve the associate homogeneous equation

y''' - y'' + y' - y = 0

which is solved by finding the roots to the characteristic equation

r^{3} - r^{2} + r - 1 = (r - 1)(r^{2} + 1) = 0

which lead to the solution of the homogeneous equation

y = c_{1}y_{1}(t) + c_{2}y_{2}(t) + c_{3}y_{3}(t) = c_{1}e^{t} + c_{2}cos(t) + c_{3}sin(t).

The Wronskian, W(t) = W(y_{1},y_{2},y_{3})(t), turns out to be

W(t) = 2e^{t},

and, defining the determinant W_{k} to be W with the k^{th} column replaced by (0,0,...,0,1), you will find that the solution (via Cramer's Rule and one integration) looks like

Y(t) = ∑_{k=1}^{3}( y_{k}(t) ∫_{t}_{0}^{t} ( (g(s)W_{k}(s)) / W(s) ) ds.

From this formula (which is essentially the result of the Method of Variation of Parameters), and computing the W_{k}'s on your own, along with W, I think you should be able to reproduce the answer given. If not, let us know where you are getting hung up, and what you've tried so far.

Best of luck, and I hope this helps!

Regards,

Hassan H.