I was solving but the coefficients based on my solution are
4
-8/3
-4
-28/15
and fill them in the blanks respectively.
Additional explanation:
The power series is the expansion of increasing power of the x variable with coefficients.
Since we are provided with y = 8x + ___x² + ___x³ + ___x⁴ + ___x⁵, we are going to represent the blanks (which are coefficients) into letters as A, B, C and D and then we solve for these unknowns. So
y = 8x + Ax² + Bx³+ Cx⁴ + Dx⁵
We differentiate the equation accordingly:
y' = 8 + 2Ax + 3Bx² + 4Cx³+ 5Dx⁴
y'' = 2A + 6Bx + 12Cx² + 20Dx³
Substituting these three equations to the differential equation becomes
(x² - x + 1)y'' - y' + 4y = 0
(x² - x + 1)(2A + 6Bx + 12Cx² + 20Dx³) - (8 + 2Ax + 3Bx² + 4Cx³+ 5Dx⁴) + 4(8x + Ax² + Bx³ + Cx⁴ + Dx⁵) = 0
2Ax² + 6Bx³ + 12Cx⁴ - 2Ax - 6Bx² - 12Cx³ + 2A + 6Bx + 12Cx² - 8 - 2Ax - 3Bx² - 4Cx³ + 32x + 4Ax² + 4Bx³ + 4Cx⁴ = 0
Seems that this equation is too long. So I did not add the like terms at this point because it is risky. Next, collect the terms that have same power of x because this is where we are going to obtain the value of unknown coefficients.
For x⁰:
2A - 8 = 0
A = 4
For x¹:
-2Ax + 6Bx - 2Ax + 32x = 0
-2A + 6B - 2A + 32 = 0
Since A = 4, so
B = -8/3
For x²:
2Ax² - 6Bx² + 12Cx² - 3Bx² + 4Ax² = 0
2A - 6B + 12C - 3B + 4A = 0
C = -4
For x³:
6B - 12C - 4C + 4B + 20D = 0
D = -28/15
There are other ways to solve this problem such as the summation of power series or integration; however these are complicated to use.
