Answer:
Step-by-step explanation:
Answer:
y = C1x^4 + (1/x²)[C2cos(√2)lnx + C3sin(√2)lnx]
Step-by-step explanation:
Given the differential equation
x³y'''(x) + 3x²y''(x) - 9xy'(x) - 24y(x) = 0 .........................................................(1)
We are required to solve the differential equation, using the substitution,
y = x^r
Differentiating this three times in succession, to obtain y', y'', and y'''. We have
y' = rx^(r - 1)
y'' = r(r - 1)x^(r - 2)
y''' = r(r - 1)(r - 2)x^(r - 3)
Using these in (1), we have
x³r(r - 1)(r - 2)x^(r - 3) + 3x²r(r - 1)x^(r - 2) - 9xrx^(r - 1) - 24x^r = 0
[r(r - 1)(r - 2) + 3r(r - 1) - 9r - 24]x^r = 0
But x^r ≠ 0
So
r(r - 1)(r - 2) + 3r(r - 1) - 9r - 24 = 0
r³ - 3r² + 2r + 3r² - 3r - 9r - 24 = 0
r³ - 10r - 24 = 0
(r - 4)(r² + 4r + 6) = 0
r - 4 = 0
r = 4 ..........................................(a)
r² + 4r + 6 = 0
Using the quadratic formula
r = [-b ± √(b² - 4ac)]/2a
Where a = 1, b = 4, and c = 6
r = [-4 ± √(16 - 24)]/2
= -4/2 ± (1/2)√(-8)
r = -2 ± i√2 ....................................(b)
From (a) and (b), we have
r = 4, -2 ± i√2
Since y = x^r
The solution is therefore
y = C1x^4 + (1/x²)[C2cos(√2)lnx + C3sin(√2)lnx]
