Write this as:
y""-8y'"+24y"-32y'+16y=2x²+1.
First solve y""-8y'"+24y"-32y'+16y=0 (homogeneous part represented by y[H]).
The characteristic equation can be written:
(r⁴-8r³+24r²-32r+16)=(r-2)⁴=0 giving us the homogeneous solution:
y=pe²ˣ where p(x)=A+Bx+Cx²+Dx³, where A, B, C, D are arbitrary constants, because r=2 is the unique solution to the characteristic equation, implying e²ˣ as the exponential factor.
y'=e²ˣ(2p+p'),
y"=e²ˣ(4p+4p'+p"),
y'"=e²ˣ(8p+12p'+6p"+p'"),
y""=e²ˣ(16p+32p'+24p"+8p'"+p"").
If these are substituted into y""-8y'"+24y"-32y'+16y the result is zero.
Therefore y[H]=e²ˣ(A+Bx+Cx²+Dx³).
Now we need to find the particular solution, that is, the right-hand side 2x²+1.
Assume that the particular solution has the form y=ax²+bx+c, where a, b, c are constants to be found (they are not arbitrary), and y[P] represents the particular solution.
For y[P]:
y'=2ax+b, y"=2a, y'"=0=y"".
Substituting these in the original DE:
24y"-32y'+16y≡2x²+1, because higher derivatives are zero.
Note the use of the identity equivalence. This implies that we have to match coefficients of powers of x on each side of the equation.
24(2a)-32(2ax+b)+16(ax²+bx+c)≡2x²+1.
Therefore, equating x² coefficients, 16a=2, a=⅛.
Equating x coefficients, -64a+16b=0, b=4a=½.
And equating constants:
48a-32b+16c=1,
3a-2b+c=1/16,
c=2b-3a+1/16=1-⅜+1/16=11/16.
So y[P]=x²/8+x/2+11/16.
The final solution is y=y[H]+y[P], that is:
y=e²ˣ(A+Bx+Cx²+Dx³)+x²/8+x/2+11/16.