Find y as a function of x if y"' + 9y' = 0, y (0) = 6, y' (0) = 9, y" (0) = 18. y (x) =

y (x) = 8 - 2cos 3x + 3sin 3x

Step-by-step explanation:

Given the equation:

y''' + 9y' = 0

The characteristic equation:

⇒r³ + 9r = 0

Solving for r, we get r = 0 and r = ± 3i

The general equation for such equation is:

y = C₁ + C₂cos 3x + C₃sin 3x

Given:

y (0) = 6

Thus, Applying in the above equation, we get

6 = C₁ + C₂ ... 1

Differentiating y, we get:

y' = - 3C₂sin 3x + 3C₃cos 3x

Given:

y' (0) = 9

Thus, Applying in the above equation, we get

9 = 3C₃

or,

C₃ = 3

Differentiating y', we get:

y'' = - 9C₂cos 3x - 9C₃sin 3x

Given:

y'' (0) = 18

Thus, Applying in the above equation, we get

18 = - 9C₂

or,

C₂ = - 2

Applying in equation 1, we get:

C₁ = 8

Thus,

y (x) = 8 - 2cos 3x + 3sin 3x