Question on variation of parameters - ODE

[cue928]

I am working on a problem requiring variation of parameters. When I calculated the wronskian, I got an answer, which differed from the book only by a "-" (mine was -, the book's was +). So I switched my functions for y1 and y2 and got the answer the book had. Is there a standard for which should be which or does it work itself out either way through the course of the problem, once you start integrating?

 

[Kreizhn]

I may be wrong, but your definition of y1 and y2 shouldn't matter. Because the Wronskian is given by a determinant, we know that switching y1 and y2 corresponds to switching a column and hence flips the sign of the Wronskian.

However, at the same time the formulas for the "coefficients" A(x) y1 + B(x) y2 also switch signs. This means that the answer should be the same regardless of which indexing scheme you give.

 

[Kreizhn]

To be more precise, let's say that the book calculates the wrongskian to be W(x). Then the coefficients will be
A(x) = -\int \frac1W y_2(x) b(x) \ dx, \qquad B(x) = \int \frac1W y_1(x) b(x) \ dx
where b(x) is the inhomogeneous factor. This gives
A(x) y_1(x) + B(x) y_2(x)

Now if you calculate -W(x), and set \hat y_1 = y_2, \hat y_2 = y_1 then
\hat A(x) = -\int \frac1{-W} \hat y_2(x) b(x) \ dx, \qquad \hat B(x) = \int \frac1{-W} \hat y_1(x) b(x) \ dx
And your solution will be
\hat A(x) \hat y_1(x) + \hat B(x) \hat y_2(x)
but you can easily calculate \hat A(x) = B(x), \hat B(x) = A(x) so your solution is equivalent to the first.

 

[cue928]

Yeah, I went ahead and ran through the two scenarios and got the same answer. I guess there could be two different answers on the wronskian that would produce the same answer. Haven't had linear algebra yet, but I suspect if I ever take it, I'll get some more insight. Thanks for your reply.