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

A, D and E is your answer

Good luck!! :) :p