If I remember correctly; you first calculate the homogeneous solution. And then the particular solution for the non-homogeneous part.
Deviding both terms by the DE becomes...
(1)
First step is the solution of 'incomplete equation' ...
(2)
... that is of the type...
(3)
A DE like (3) can often be solved in terms of an auxiliary function defined as...
(4)
Deriving both terms of (4) we obtain first ...
(5)
... deriving again...
(6)
... and then with two more steps...
(7)
Taking into account (7) the (3) can be written in terms of as...
(8)
... where...
(9)
A little pause before proceeding...
Kind regards
In the last post we have 'attacked' the 'incomplete' DE...
(1)
... swapping the y with the u variable defined as...
(2)
... and we are arrived to the DE...
(3)
... where...
(4)
In our case we have...
, (5)
... so that...
, (6)
The (6) shows us that the solution of (1) in terms of is relatively 'easy' and is...
(7)
... so that is...
(8)
The solution of the 'complete' DE is left for further study...
Kind regards
Continuing with Chisigma's solution, then I'll finish it if that's alright. I'll use variation of parameter:
Then and therefore
Let
Substituting that into the original DE we obtain the following two equations to solve:
Then
Substituting that one into the second one, I get or so that and in which we drop the arbitrary constants in this case since we're only looking for a particular solution to add to the complimentary solution.
Therefore, the general solution is
or: