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 :
<u>y = C₁ + C₂cos 3x + C₃sin 3x</u>
Given:
y(0) = 6
Thus, Applying in the above equation, we get
<u>6 = C₁ + C₂ .......................................................1</u>
Differentiating y , we get:
<u>y' = -3C₂sin 3x + 3C₃cos 3x</u>
Given:
y'(0) = 9
Thus, Applying in the above equation, we get
<u>9 = 3C</u><u>₃</u>
or,
<u>C</u><u>₃</u><u> = 3</u>
Differentiating y' , we get:
<u>y'' = -9C₂cos 3x - 9C₃sin 3x</u>
Given:
y''(0) = 18
Thus, Applying in the above equation, we get
<u>18 = -9C</u><u>₂</u>
or,
<u>C</u><u>₂</u><u> = -2</u>
Applying in equation 1 , we get:
<u>C₁ = 8</u>
Thus,
<u>y(x) = 8 - 2cos 3x + 3sin 3x</u>
