If we write the DE in terms of LT we have...
... and from it we obtain...
The solution of the DE can be [formally...] obtained computing the inverse LT of (2) and is...
Variation of parameters:
The general solution to the associated homogeneous equation, y"- y= 0, is, as you say, .
In "variation of parameters" we look for a solution of the form . Note that there are an infinite number of such solutions. In fact, given any solution, Y(x), we could just take u(x) identically 0 and then solve for v(x) or, conversely, take v(x) identically 0 and solve for u(x).
In any case, if then . Since there are an infinite number of "u" and "v" that will work, we can "narrow the search" and simplify by requiring, also, that .
With that we have and, differentiating again, . Putting that into the differential equation, .
Our requirement that " means that we will have no u" or v" and the fact that and satisfy the homogeneous equation means there is no u or v. We have only u' and v' in the equation.
Now we can solve the two equations and "algebraically" for u' and v'. Adding the two equations gives so . Subtracting the second equation from the first gives so .
Integrate to find u(x) and v(x). You can do those integrals by integration by parts, taking and in the one case and in the other. That will reduce to an integral in . Repeat until you have completely reduced that to . After a few example, you should be able to find a formula for the general case.
I will confess that I would have done this a third way, using "undetermined coefficients". Try a specific solution of the form , put that into the differential equation and solve for the coefficients.
thanks a lot HallsofIvy for the post...and certainly the part solved is almost same as i did except the last part...and yes i am trying to do the integral...its like series of n n-1 n-2 n-3 and in the end for the power of is n!
so t sum up all the series to find a general solution ....
will upload the result