The characteristic solution follows from solving the characteristic equation,

so that

A guess for the particular solution may be

, but this is already contained within the characteristic solution. We require a set of linearly independent solutions, so we can look to

which has second derivative

Substituting into the ODE, you have



Therefore the particular solution is

Note that you could have made a more precise guess of

but, of course, any solution of the form

is already accounted for within

.
Using polynomial long division, we get
3x^3+6x^2+11x
_____________
(x+2) | 3x^4-x^2+cx-2
-(3x^4+6x^3)
____________
6x^3-x^2+cx-2
- (6x^3+12x^2)
_____________
11x^2+cx-2
-(11x^2+22x)
__________
(22+c)x-2.
If you're wondering how I did the long division, what I essentially did was get the first value (at the start, it was 3x^4) and divided it by the first value of the divisor (which in x+2 was x) to get 3x^3 in our example. I then subtracted the polynomial by the whole divisor multiplied by, for example, 3x^3 and repeated the process.
For this to be a perfect factor, (x+2)*something must be equal to (22+c)x-2. Focusing on how to cancel out the 2, we have to add 2 to it. To add 2 to it, we have to multiply (x+2) by 1. However, there's a catch, which is that we subtract whatever we multiply (x+2) by, so we have to multiply it by -1 instead. We still need to cross out (22+c)x. Multiplying (x+2) by -1, we get
(-x-2) but by subtracting the whole thing from something means that we have to add -(-x-2)=x+2 to something to get 0. x+2-x-2=0, xo (22+c)x-2 must equal -x-2, meaning that (22+c)=-1 and c=-23
Answer:
the answer is A
Step-by-step explanation:
cross multiply 16×4 =64