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Radda [10]
3 years ago
5

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

Mathematics
1 answer:
Andreas93 [3]3 years ago
4 0

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>

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