TITLE: How to solve $(x^2+1)y''-2xy'+2y=6(x^2+1)^2$, $y_1=x$
QUESTION [0 upvotes]: I've doing this exercise, but I'm not sure if I'm right:
The first thing I did was to solve the homogeneous equation:
$$y''-2\frac x{x^2+1}y'+\frac2{x^2+1}y =0$$
Then I used the formula to find $y_2$:
$$y_2= y_1\int \frac{e^{-\int p\mathop{dx}}}{y_1^2}\mathop{dx},$$
and the answer I got was $y_2= -1 -x\arctan(x)$. So the answer to the homogeneous equation was:
$$y= c_1(x) + c_2(-1-x\arctan(x)).$$
Then I passed to solve the nonhomogeneous differential equation, but the answer I got it's too large.
Can anyone please help me or tell me if I'm doing well?

REPLY [1 votes]: If you know the solution of homogeneous equation, to solve nonhomogeneous equation you only need to find one particular solution.
One can see that on the left you have a polynomial of 4-th degree. If $y(x)$ is a polynomial too, then all three terms on the right have the same degree. So it makes sense to search for the solution within polynomials of 4th degree.
If you know that a polynomial has form
$$
y_n(x) = ax^4+bx^3+cx^2+dx +e,
$$
can you find at least one set of $a,b,c,d,e$?
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