Question

Use the substitution y equals x Superscript r to find a general solution to the given equation for x greater than 0. x cubed y triple prime (x )plus 3 x squared y double prime (x )minus 9 xy prime (x )minus 24 y (x )equals0

Answer 1

Answer:

The problem solution  is given in the attachments.

Answer 2

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]