Okay, it is easy to check that y1 and y2 really are independent solutions to the homogeneous equation (and you should always check).

We look for a solution to the entire equation of the form . Then . There are, in fact, an infinite number of such functions, u(x) and v(x), (given a solution y, pickanyfunction to be u(x) and solve for v), we can simplify, and narrow the search, by requiring that . That reduces y' to so and the equation becomes

Our clever requirement that means that there are no u" or v" terms in the equation and the fact that y1 and y2 satisfy the homogeneous equation means that there are no u or v terms! We have, now, two equations, and to solve for u' and v'. For the first, we can divide through by x-1 to get . Solve those equations, algebraically, for u' and v', then integrate to find u and v.