Answer:
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
